Question

You observe someone pulling a block of mass 41 kg across a low-friction surface. While they pull a distance of 5 m in the direction of motion, the speed of the block changes from 5 m/s to 6 m/s. Calculate the magnitude of the force exerted by the person on the block.

Answer 1

Answer: 45.1N

Explanation:

Answer: 45.1N

Explanation:

The work done in the block is equal to change in energy is the system. We then go ahead to calculate W with the given values of initial and final speed.

W= Ef - Ei

W= 1/2MVf² - 1/2MVi²

W= 1/2M (Vf² - Vi²)

W= 1/2 * 41 (6²-5²)

W= 1/2 * 41 * 11

W= 225.5J

We note the definition of work and solve for F.

Taking into cognizance the displacement we have been given in the question. We then have

F = W/x

F = 225.5/5

F = 45.1N

Answer 2

Answer:

45.1 N

Explanation:

Using Newton's Fundamental equation of Kinematics,

F = ma............. Equation 1

Where F = force exerted by the person on the block, m = mass of the block, a = acceleration of the block

We can look for the acceleration of the block by applying newton's equation of motion,

Using,

v² = u²+2as....................... Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation

a = (v²-u²)/2s................... Equation 3

Given: v = 6 m/s, u = 5 m/s, s = 5 m

Substitute into equation 3

a = (6²-5²)/(2×5)

a = 11/10

a = 1.1 m/s²

Also Given: m = 41 kg

Substitute into equation 1

F = 41(1.1)

F = 45.1 N