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.

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

asuwafohjames asuwafohjames · Answer 2 · 2020-03-17T04:19:54+01:00

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