Question

A 6 kg sled is initially at rest on a horizontal road. The sled is pulled a distance of 2.8 m by a force of 36 N applied to the sled at an angle of 30o to the horizontal. Find the change in the kinetic energy of the sled.

Answer 1

The horizontal component of the applied force is:

Fx = F*Cos(theta) = 36*Cos(30) = 31.1 N

The vertical component of the applied force is:

Fy = F*Sin(theta) = 36*Sin(30) = 18 N

Since the Fy is less than the weight of the sled (m*g), the sled stays on the ground.

The work done is the force multiplied by the distance of exertion. In this case, there is only motion in the horizontal, so use the horizontal component, multiplied by the distance pulled:

W = Fx*d = 31.1*2.8 = 87.08 J

The force of friction is:

F_f = ?*N

The normal force is the weight of the sled (m*g) minus the applied vertical component of force:

N = m*g - Fy = 6*9.8 - 18 = 40.8 N

F_f= ?*N = 0.4*40.8 = 16.32 N

The energy dissipated by friction is equal to the work done against friction:

W_f = F_f*d = 16.32*2.8 = 45.69 J

The Kinetic Energy (KE) of the sled is the difference between the applied work, and the work lost to friction:

KE = 87.08 - 45.69 = 41.38 J