Sally applies a total force of 379N with a rope to drag a wooden crate of mass 1

Question

Sally applies a total force of 379N with a rope to drag a wooden crate of mass 100kg across a horizontal floor over distance of 45m. The rope tied to the crate is pulled at an angle of 60 degrees relative to the floor as shown. The coefficient of kinetic friction between the floor and the crate is muk = 0.2.

A.) How much work is done by the FRICTION FORCE?

B.) How much work is done by the GRAVITY FORCE?

C.) How much work is done by the NORMAL FORCE?

D.) If the crate started from rest, what is its speed at the end of the 45m displacement?

Explanation / Answer

In vertical balancing force,

N + Fsin60 = mg

N + ( 379 x sin60) = 100 x 9.81

N = 652.78 N

friction force f = uN = 0.2 x 652.78 = 130.56 N

A) Work done by friction = f.d = - 130.56 x 45cos180 = - 5874.99 N

b) gravity force makes 90 deg with displacement.

W = mgdcos90 = 0

c) normal force also makes 90 deg angle.

so work done = 0

D) using work energy theorem,

Work done by force + work by friction + work done by gravity + work done normal = change in KE

(379 x 45 x cos60) + ( - 5875) + 0 + 0 = 100v^2 /2 - 0

v = 7.28 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.