Question

A 5.0-kg object is pulled along a horizontal surface at a constant speed by an 11.5 N force acting 30degree above the horizontal. How much work is done by this force as the object moves 7.4 m? 78 j 82 J 85 J 74 J 43 j A trunk is pulled by a constant horizontal force of 12 N at an angle of 15degree above the horizontal. The work done by the applied force after the trunk has move 4.15 m is -44.0 j -48.0j +44.0 j +48.0 j -50.0 j A trunk is pulled by a horizontal force of 15 N at an angle of 30degree above the horizontal. If the trunk weighs 137.5 N the coefficient of kinetic friction is 0.10 0.12 0.15 0.18 0.20 A 2.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N at acting at angle of 37degree above the horizontal as shown. The block is displaced 7.0 m. Calculate the work d by the friction force during this displacement. [mu_k = 0.50]

Answer 1

1. W = F.d = |F||d| cos@

@ is the angle between F and d.

W = 11.5 * 7.4 * cos30 = 73.7 J
ANs(a)

2. W = 12 * 4.15 * cos15 = 48.1 J

ans(D)

4. In vertical,

N + Fsin30 - mg = 0

N = 137.5 - (15 sin30) = 130 N

in horizontal,

f - Fcos30 = 0

f = friction force = uk N

u N = F cos30

130 uk = 15 cos30

uk = 0.10

4. in vertical,

Fnet= N + 16sin37 - mg = 0

N = (2 x 9.8) - (16 sin37) = 9.97 N

friction force, f =uk N = 0.50 x 9.97 = 5 N

work done = f.dcos180 = 5 * 7 *cos180 = - 35 J