A box of mass 20 kg slides from rest at the top of the incline shown to the bott
ID: 1290261 • Letter: A
Question
A box of mass 20 kg slides from rest at the top of the incline shown to the bottom where it is 3 m vertically below and 12 m horizontally to the right of its initial position. If the block's speed at the bottom is 6 m/s what was the work done on the box by friction?
(A) +230 J
(B) -130 J
(C) +130 J
(D) -230 J
(E) +950 J
Please show work and explain!
A box of mass 20 kg slides from rest at the top of the incline shown to the bottom where it is 3 m vertically below and 12 m horizontally to the right of its initial position. If the block's speed at the bottom is 6 m/s what was the work done on the box by friction? (A) +230 J (B) -130 J (C) +130 J (D) -230 J (E) +950 JExplanation / Answer
Use the law of conservation of energy to find the work done by the friction.
mgh + Wk = 0.5mv2
Hence, the work done by the friction is,
Wk = 0.5mv2 - mgh
= 0.5(20 kg)(6 m/s)2 - (20 kg)(9.8 m/s2)(3 m)
= -228 J
= -230 J (nearly)
Answer: option (D)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.