Question

3. A box having a mass equal to 10.0 kg is initially at rest on a horizontal, frictionless floor. A pulling force having a magnitude of 5.00 N is applied at an angle of 30.0° above the horizontal. The box is pulled a distance of 15.0 m. 7 ] = 5.00 N 30.0° (a) What is the work done by the pulling force? (b) What is the work done by gravity? (c) What is the work done by the normal force? (d) What is the total work done on the box? (e) What is the speed of the box after being pulled 15.0 m?

Answer 1

a)

P = pulling force = 5.00 N

d = displacement of the box = 15.0 m

$\theta$ = angle between the pulling force and displacement = 30

Work done by the pulling force is given as

W_p = P d Cos$\theta$ = (5.00) (15.0) Cos30

W_p = 64.95 J

b)

m = mass of the box = 10 kg

F_g = force of gravity on the box = mg = 10 x 9.8 = 98 N

d = displacement of the box = 15.0 m

$\theta$_g = angle between the force of gravity and displacement = 90

Work done by the force of gravity is given as

W_g = F_g d Cos$\theta$_g = (98) (15.0) Cos90

W_g = 0 J

c)

m = mass of the box = 10 kg

F_n = normal force on the box = mg - P Sin30 = 98 - 5 Sin30 = 95.5 N

d = displacement of the box = 15.0 m

$\theta$_n = angle between the normal force and displacement = 90

Work done by the normal force is given as

W_n = F_n d Cos$\theta$_n = (95.5) (15.0) Cos90

W_n = 0 J

d)

W_net = total work done on the box

Total work done on the box is given as

W_net = W_p + W_g + W_n

W_net = 64.95 + 0 + 0

W_net = 64.95 J

e)

v = speed of the box

Using work-change in kinetic energy equation

Change in kinetic energy = Net work done

(0.5) m v² = W_net

(0.5) (10) v² = 64.95

v = 3.6 m/s