Question

A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘ below the horizontal.
a. Find the work done by the force F⃗ P on the cart if the ramp is 6.5 m long.
b. Find the work done by the force mg⃗ on the cart.

Answer 1

Answer:

(a)The work done by the force Fp is 198.6J

(b) The work done by the weight mg is -198.6J

The net force acting on the cart is zero. This is because the cart is moving with a constant velocity and by newton's first law the the net force on the cart is equal to zero.

Fp was calculated to be equal to 34.94 N.

Explanation:

In order to solve this kind of problems successfully, the best approach is to resolve all forces acting on the cart parallel and perpendicular to the ramp surface. So that the x-axis is parallel to the ramp surface and the y axis is perpendicular to the ramp surface.

Fpx = FpCos 29°

Fpy = FpSin 29°

Wx = mg sin12°

Wy = mg cos 12°

Summation Fx = 0

And Summation Fy = 0

The full solution can be found below in the attachment.

Thank you for reading and I hope this is helpful to you.

Answer 2

Answer: a. 198.6J b. - 198.6J

Explanation: Parameters given:

m = 15kg

g = 9.8m/s²

∅ = 12°

a. Work done by the force Fp on the cart if the ramp is 6.5m long.

Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N

Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J

b. The work done by the force mg on the cart.

Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance

= 15kg x -9.8m/s² x Sin12° x 6.5m

= - 198.6J