https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://ejercicios-fyq.com/Force-a-man-must-exert-to-pull-a-sled-8334 https://ejercicios-fyq.com/Force-a-man-must-exert-to-pull-a-sled-8334 2024-10-23T03:01:15Z text/html es F_y_Q Dynamics SOLVED Friction force Newton's second law <p>A man pulls a sled up a ramp using a rope attached to the front, as illustrated in the figure. <br class='autobr' /> The sled has a mass of 80 kg. The kinetic friction coefficient between the sled and the ramp is (), the angle between the ramp and the horizontal is , and the angle between the rope and the ramp is . What force must the man exert to keep the sled moving at a constant speed?</p> - <a href="https://ejercicios-fyq.com/Dynamics" rel="directory">Dynamics</a> / <a href="https://ejercicios-fyq.com/Dynamics-663" rel="tag">Dynamics</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a>, <a href="https://ejercicios-fyq.com/Friction-force" rel="tag">Friction force</a>, <a href="https://ejercicios-fyq.com/Newton-s-second-law" rel="tag">Newton's second law</a> <div class='rss_texte'><p>A man pulls a sled up a ramp using a rope attached to the front, as illustrated in the figure.</p> <div class='spip_document_1348 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'> <figure class="spip_doc_inner"> <img src='https://ejercicios-fyq.com/local/cache-vignettes/L295xH258/zzz-790a0.jpg?1758420408' width='295' height='258' alt='' /> </figure> </div> <p>The sled has a mass of 80 kg. The kinetic friction coefficient between the sled and the ramp is (<img src='https://ejercicios-fyq.com/local/cache-vignettes/L68xH16/64cc50cc8344bf421cebea37a6b3ed33-809ba.png?1732956198' style='vertical-align:middle;' width='68' height='16' alt="\mu_k = 0.70" title="\mu_k = 0.70" />), the angle between the ramp and the horizontal is <img src='https://ejercicios-fyq.com/local/cache-vignettes/L22xH13/f13e28ce829b262174de9bc35f8a8b7d-c9043.png?1732956198' style='vertical-align:middle;' width='22' height='13' alt="25^o" title="25^o" />, and the angle between the rope and the ramp is <img src='https://ejercicios-fyq.com/local/cache-vignettes/L32xH42/9a852d58dca44420883431ed11b2f8b2-559a4.png?1732956198' style='vertical-align:middle;' width='32' height='42' alt="35^o" title="35^o" />. What force must the man exert to keep the sled moving at a constant speed?</math></p></div> <hr /> <div <div class='rss_ps'><p>To analyze the forces acting on the sled, decompose them according to the system formed by the direction of movement and a perpendicular axis: <br/></p> <div class='spip_document_1349 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'> <figure class="spip_doc_inner"> <img src='https://ejercicios-fyq.com/IMG/jpg/z.jpg' width="268" height="228" alt='' /> </figure> </div> <p><br/> The components of the weight and applied force are: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/914aec00087a7a6616bff968b6c05f15.png' style="vertical-align:middle;" width="195" height="52" alt="\left p_x = m\cdot g\cdot sen\ 25 ^o \atop p_y = m\cdot g\cdot cos\ 25^o \right \}" title="\left p_x = m\cdot g\cdot sen\ 25 ^o \atop p_y = m\cdot g\cdot cos\ 25^o \right \}" /> <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/91d1053de8a842a0a58213a1e1f66b56.png' style="vertical-align:middle;" width="167" height="52" alt="\left F_x = F\cdot cos\ 35 ^o \atop F_y = F\cdot sen\ 35^o \right \}" title="\left F_x = F\cdot cos\ 35 ^o \atop F_y = F\cdot sen\ 35^o \right \}" /> <br/> <br/> In the direction perpendicular to the ramp, the sum of the forces must be zero. Consider the "y-component" of the weight to solve for the normal force: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/5d6c3b0c3fa28a8ac5a18a07da98c6a9.png' style="vertical-align:middle;" width="481" height="22" alt="p_y = N + F_y\ \to\ \color[RGB]{2,112,20}{\bf N = m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o}}" title="p_y = N + F_y\ \to\ \color[RGB]{2,112,20}{\bf N = m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o}}" /> <br/> <br/> In the direction of movement, the sum of the forces must also be zero because the sled moves at a constant speed. This gives: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/d08ce8c4ff274e375a8a4dbf748ed443.png' style="vertical-align:middle;" width="347" height="20" alt="F_x - p_x- F_R = 0\ \to\ \color[RGB]{2,112,20}{\bm{F_x= p_x - F_R}}" title="F_x - p_x- F_R = 0\ \to\ \color[RGB]{2,112,20}{\bm{F_x= p_x - F_R}}" /> <br/> <br/> The friction force is the product of the normal force and the coefficient of friction: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/a461c7975b0794fae49e6685e5a73b09.png' style="vertical-align:middle;" width="571" height="23" alt="F\cdot cos\ 35^o = \mu_c(m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o) + m\cdot g\cdot sen\ 25^o" title="F\cdot cos\ 35^o = \mu_c(m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o) + m\cdot g\cdot sen\ 25^o" /> <br/> <br/> Substitute the known values into the equations to solve for the force exerted by the man: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/042a3cabf1ae528823e6b293e081baf7.png' style="vertical-align:middle;" width="680" height="28" alt="0.819F = 497.4 - 0.402F + 331.3\ \to\ 1.221F = 828.7\ \to\ \fbox{\color[RGB]{192,0,0}{\bf F= 678.7\ N}}" title="0.819F = 497.4 - 0.402F + 331.3\ \to\ 1.221F = 828.7\ \to\ \fbox{\color[RGB]{192,0,0}{\bf F= 678.7\ N}}" /></p> </math></p></div> https://ejercicios-fyq.com/Acceleration-of-an-elevator-with-different-readings-on-the-weighing-scale-7068 https://ejercicios-fyq.com/Acceleration-of-an-elevator-with-different-readings-on-the-weighing-scale-7068 2021-03-12T05:53:59Z text/html es F_y_Q Acceleration Non-inertial system Dynamics SOLVED <p>A person with a mass of 70 kg stands on a weighing scale in a moving elevator and observes different readings. Determine the acceleration of the elevator and wether it is moving up, down, accelereting, decelerating or stationary if the scale readings are: a) 66 kg, b) 74 kg and c) 70 kg.</p> - <a href="https://ejercicios-fyq.com/Dynamics" rel="directory">Dynamics</a> / <a href="https://ejercicios-fyq.com/Acceleration" rel="tag">Acceleration</a>, <a href="https://ejercicios-fyq.com/Non-inertial-system" rel="tag">Non-inertial system</a>, <a href="https://ejercicios-fyq.com/Dynamics-663" rel="tag">Dynamics</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>A person with a mass of 70 kg stands on a weighing scale in a moving elevator and observes different readings. Determine the acceleration of the elevator and wether it is moving up, down, accelereting, decelerating or stationary if the scale readings are: a) 66 kg, b) 74 kg and c) 70 kg.</p></div> <hr /> <div <div class='rss_ps'><p>Weight and normal force have opposite sense and the following equation is used: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/824578b0ae81d170cf8c60d8ac58d1fb.png' style="vertical-align:middle;" width="356" height="50" alt="p - N = m\cdot a\ \to\ \color[RGB]{2,112,20}{\bm{a= \frac{(m - m^{\prime})\cdot g}{m}}}" title="p - N = m\cdot a\ \to\ \color[RGB]{2,112,20}{\bm{a= \frac{(m - m^{\prime})\cdot g}{m}}}" /> <br/> <br/> Just substitute the values and calculate the acceleration in each case: <br/> <br/> a) <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/eb9528a74984c5a0767e8db66ccdec15.png' style="vertical-align:middle;" width="339" height="60" alt="a = \frac{(70 - 66)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0.56\ \frac{m}{s^2}}}}" title="a = \frac{(70 - 66)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0.56\ \frac{m}{s^2}}}}" /></p> <br/> <b>The elevator is moving down</b>. <br/> <br/> b) <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/4c6006e382a7c44ee3f18b2ae5aaaf60.png' style="vertical-align:middle;" width="359" height="60" alt="a = \frac{(70 - 74)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{- 0.56\ \frac{m}{s^2}}}}" title="a = \frac{(70 - 74)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{- 0.56\ \frac{m}{s^2}}}}" /></p> <br/> <b>The elevator is moving up</b>. <br/> <br/> c) <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/1ec9cfef4e5b95ceeb98f835d2c9f51d.png' style="vertical-align:middle;" width="309" height="60" alt="a = \frac{(70 - 70)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0\ \frac{m}{s^2}}}}" title="a = \frac{(70 - 70)\ \cancel{kg}\cdot 9.8\ \frac{m}{s^2}}{70\ \cancel{kg}}= \fbox{\color[RGB]{192,0,0}{\bm{0\ \frac{m}{s^2}}}}" /></p> <br/> <b>The elevator is not moving</b>.</math></p></div> https://ejercicios-fyq.com/Lengthening-of-a-spring-when-a-force-is-applied-4843 https://ejercicios-fyq.com/Lengthening-of-a-spring-when-a-force-is-applied-4843 2019-11-11T18:42:07Z text/html es F_y_Q EDICO Hooke's law SOLVED <p>A spring has an elastic constant of 120 N/m. Calculate how much it will lengthen if a force of 35 N is applied.</p> - <a href="https://ejercicios-fyq.com/Dynamics" rel="directory">Dynamics</a> / <a href="https://ejercicios-fyq.com/EDICO" rel="tag">EDICO</a>, <a href="https://ejercicios-fyq.com/Hooke-s-law" rel="tag">Hooke's law</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>A spring has an elastic constant of 120 N/m. Calculate how much it will lengthen if a force of 35 N is applied.</p></div> <hr /> <div <div class='rss_ps'><p>Use the Hooke's law to solve the problem: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/ed8208e3cb3d08eebc0f94a4fe04ef30.png' style="vertical-align:middle;" width="430" height="57" alt="F = k\cdot \Delta x\ \to\ \Delta x = \frac{F}{k} = \frac{35\ \cancel{N}}{120\ \frac{\cancel{N}}{m}}= \fbox{\color[RGB]{192,0,0}{\bf 0.29\ m}}" title="F = k\cdot \Delta x\ \to\ \Delta x = \frac{F}{k} = \frac{35\ \cancel{N}}{120\ \frac{\cancel{N}}{m}}= \fbox{\color[RGB]{192,0,0}{\bf 0.29\ m}}" /></p> </math></p> <p> <br/></p> <p><b><a href="https://ejercicios-fyq.com/Situaciones-de-aprendizaje/EDICO/Ej_4843.edi" download>Download the statement and the solution of the problem in EDICO format if you need it.</a></b></p></div>