A 5.40 kg block is set intomotion up an inclined plane with an initial speed of
ID: 1764038 • Letter: A
Question
A 5.40 kg block is set intomotion up an inclined plane with an initial speed ofv0 = 8.10 m/s. Theblock comes to rest after traveling 3.00 m along the plane, whichis inclined at an angle of 30.0° to the horizontal. (a) For this motion, determine the change inthe block's kinetic energy.J
(b) For this motion, determine the change in potential energy ofthe block-Earth system.
J
(c) Determine the frictional force exerted on the block (assumed tobe constant).
N
(d) What is the coefficient of kinetic friction?
A 5.40 kg block is set intomotion up an inclined plane with an initial speed ofv0 = 8.10 m/s. Theblock comes to rest after traveling 3.00 m along the plane, whichis inclined at an angle of 30.0° to the horizontal. (a) For this motion, determine the change inthe block's kinetic energy.
J
(b) For this motion, determine the change in potential energy ofthe block-Earth system.
J
(c) Determine the frictional force exerted on the block (assumed tobe constant).
N
(d) What is the coefficient of kinetic friction?
(a) For this motion, determine the change inthe block's kinetic energy.
J
(b) For this motion, determine the change in potential energy ofthe block-Earth system.
J
(c) Determine the frictional force exerted on the block (assumed tobe constant).
N
(d) What is the coefficient of kinetic friction?
Explanation / Answer
(a) Change in KE = final KE - initial KE = 0 - (1/2) mv2 = . = - (1/2) * 5.40 *8.102 = - 177.15 Joules . (b) Change in PE = mgh = 5.40 *9.80 * 3.00sin30 = 79.38Joules . (c) total work by friction = total change in energy . -friction force * distance = change inKE + change in PE . -friction force * 3.00 = -177.15 + 79.38 . friction force = 32.59 Newtons (the answershould be positive, but if not try negative) . (d) friction force = coeff *normal force . 32.59 = coeff * mgcos . 32.59 = coeff * 5.40 * 9.80 * cos30 . coeff = 0.711Related Questions
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