Question

A boy pulls a 5-kg cart with a 20-N force at an angle of 30° above the horizontal for a length of time. Over this time frame, the cart moves a distance of 12 m on the horizontal floor. (a) Find the work done on the cart by the boy. (b) What will be the work done by the boy if he pulled with the same force horizontally instead of at an angle of 30° above the horizontal over the same distance?

a) 120 J
b) 207 J
c) 240 J
d) 360 J

Answer 1

The work done on the cart by the boy is 240 J in both cases: when he pulls at an angle of 30° above the horizontal and when he pulls horizontally.

Step-by-step explanation:

To solve this problem, we need to use the formula for work:

Work (W) = Force (F) x Distance (d) x cos(theta)

(a) For the first part of the question, the force applied by the boy is 20 N, the distance is 12 m, and the angle is 30° above the horizontal. We can substitute these values into the formula:

W = 20 N x 12 m x cos(30°)

W = 240 J

Therefore, the work done on the cart by the boy is 240 J.

(b) For the second part of the question, the force applied by the boy is still 20 N, but now the angle is 0° (horizontal). The distance is still 12 m. We can again substitute these values into the formula:

W = 20 N x 12 m x cos(0°)

W = 20 N x 12 m x 1

W = 240 J

Therefore, the work done on the cart by the boy if he pulled horizontally is also 240 J.