The net external force on the 24-kg mower is stated to be 51 n. if the force of friction opposing the motion is 24 n, what force f (in newtons) is the person exerting on the mower? suppose the mower is moving at 1.5 m/s when the force f is removed. how far will the mower go before stopping?

a. The force being exerted on the mower by the person is 75 Newton.

b. The distance covered by the mower before stopping is 1.13 meters.

Given the following data:

To find the force and distance covered by the mower before stopping;

First of all, we would determine the force being applied by the person:

Where:

Substituting the values, we have:

Next, we would determine the acceleration of the mower:

Since the force (F) applied by the person is removed, only the force of friction would act on the mower.

Acceleration = -1

To find the distance, we would use the third equation of motion:

S = 1.13 meters.

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PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

now we also know that friction force due to ground is given as

now we have

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

now we can find the distance by kinematics

so the distance moved will be 1.125 m

i think it would be 5