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D) Now suppose that the mass of the box is 5 kg and there is friction present on the ramp as it slides down, slowing its descent. If its specd on reaching the bottom of the ramp is 5 m/s, how much total work is done by the net force on the box as it slides from the top to the bottom? E) How much work is done by the force of gravity on the box as it slides from the top to the bottom? F) How much work is done by the force of friction on the box as it slides from the top to the bottom?

D) Now suppose that the mass of the box is 5 kg and there is friction present on the ramp as it slides down, slowing its descent. If its specd on reaching the bottom of the ramp is 5 m/s, how much total work is done by the net force on the box as it slides from the top to the bottom? E) How much work is done by the force of gravity on the box as it slides from the top to the bottom? F) How much work is done by the force of friction on the box as it slides from the top to the bottom?

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Parts D, E, F and G. I’ve already done A B and C
1:09 PM Sun Apr 10 *e 81% +: 0 A) 53.2 6 m O 348 368 8 m FaT ル° " ngeosa mg sind mg Fuer" Nja(-mgcos8;)• (-mgsin8i). FaET, -ngsind i as -gsind a= -9.8m/s2. Sın (368) a 5.87 m/s² (left/down inc B) v v² - v,² = 2ah 2+ 2ah • 0= Lah V¢= /2ah V 12-5.87/32.l0m V 17.6" V:10.84 m/s へ 15 c) W= AK W= Fd ☆ N= (mqsine)d
Transcribed Image Text:1:09 PM Sun Apr 10 *e 81% +: 0 A) 53.2 6 m O 348 368 8 m FaT ル° " ngeosa mg sind mg Fuer" Nja(-mgcos8;)• (-mgsin8i). FaET, -ngsind i as -gsind a= -9.8m/s2. Sın (368) a 5.87 m/s² (left/down inc B) v v² - v,² = 2ah 2+ 2ah • 0= Lah V¢= /2ah V 12-5.87/32.l0m V 17.6" V:10.84 m/s へ 15 c) W= AK W= Fd ☆ N= (mqsine)d
1:09 PM Sun Apr 10 ... 3) 3. A box, initially at rest, is placed on the top of the frictionless incline shown. A) Using a free body diagram and Newton's second law, find the box's acceleration in m/s'. B) Use the constant acceleration equations to find the speed of the box when in reaches the bottom of the incline. Find the work done by the constant net force on the box as it slides from the top of the incline to the bottom. Use the work-kinetic energy theorem to find the speed of the box when it reaches the bottom of the incline and check that your answer agrees with B). D) Now suppose that the mass of the box is 5 kg and there is friction present on the ramp as it slides down, slowing its descent. If its spced on reaching the bottom of the ramp is 5 m/s, how much total work is done by the net force on the box as it slides from the top to the bottom? E) How much work is done by the force of gravity on the box as it slides from the top to the bottom? F) How much work is donc by the force of friction on the box as it slides from the top to the bottom? G) If the frictional force is constant as it slides down, what is its magnitude in Newtons? A) 53.2 arctano= 6 8 10 6 m O: 36.8° 36.8° 8 m FAET 532 36.8 mgcoss ma -mgsind T リ、
Transcribed Image Text:1:09 PM Sun Apr 10 ... 3) 3. A box, initially at rest, is placed on the top of the frictionless incline shown. A) Using a free body diagram and Newton's second law, find the box's acceleration in m/s'. B) Use the constant acceleration equations to find the speed of the box when in reaches the bottom of the incline. Find the work done by the constant net force on the box as it slides from the top of the incline to the bottom. Use the work-kinetic energy theorem to find the speed of the box when it reaches the bottom of the incline and check that your answer agrees with B). D) Now suppose that the mass of the box is 5 kg and there is friction present on the ramp as it slides down, slowing its descent. If its spced on reaching the bottom of the ramp is 5 m/s, how much total work is done by the net force on the box as it slides from the top to the bottom? E) How much work is done by the force of gravity on the box as it slides from the top to the bottom? F) How much work is donc by the force of friction on the box as it slides from the top to the bottom? G) If the frictional force is constant as it slides down, what is its magnitude in Newtons? A) 53.2 arctano= 6 8 10 6 m O: 36.8° 36.8° 8 m FAET 532 36.8 mgcoss ma -mgsind T リ、
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