Question

1. A 7.00 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 2.40 m by a force of 18.0 N applied to the sled at an angle of 24° to the horizontal. Find the change in the kinetic energy of the sled.

2. A 0.24-kg stone is thrown vertically upward with an initial velocity of 7.10m/s from a height 1.30 m above the ground. What is the potential energy of the stone at its maximum height relative to the ground?

Answer 1

1.

F = force applied = 18 N

d = displacement = 2.40 m

= angle of the force with the displacement = 24

W = work done on the sled by the applied force

KE = change in kinetic energy

Work done on the sled by the applied force

W = F d Cos

W = (18) (2.40) Cos24

W = 39.5 J

Using work-change in kinetic energy theorem

Change in kinetic energy = work done

KE = W

KE = 39.5 J

2.

h_i = height of the stone at the time of launch = 1.30 m

v_i = initial speed of the stone at the time of launch = 7.10 m/s

h_f = height of the stone at the highest point = ?

v_f = final speed of the stone at the highest point = 0 m/s

m = mass of stone

Using conservation of energy

Total energy at the point of launch = Total energy at the highest point

KE_i + PE_i = KE_f + PE_f

(0.5) m v_i² + mgh_i = (0.5) m v_f² + mgh_f

(0.5) (0.24) (7.10)² + (0.24) (9.8) (1.30) = (0.5) (0.24) (0)² + (0.24) (9.8) h_f

h_f = 3.87 m