Question

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?

Answer 1

Answer:

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

$F_{net} = 51 N$

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

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

Answer 2

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:

• Net external force = 51 Newton

• Force of friction = 24 Newton

• Mass of mower = 24 kg

• Initial velocity = 1.5 m/s

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:

$Net\;force = F - F_f$

Where:

• $F_f$ is the force of friction opposing the motion.

• $F$ is the force being applied by the person.

Substituting the values, we have:

$51 = F - 24\\\\F = 51 + 24\\\\F = 75 \;Newton$

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.

$Acceleration = \frac{-F_f}{mass} \\\\Acceleration = \frac{-24}{24}$

Acceleration = -1 $m/s^2$

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

$V^2 = U^2 + 2aS\\\\0^2 = 1.5^2 + 2(-1)S\\\\0 = 2.25 - 2S\\\\2S = 2.25\\\\S = \frac{2.25}{2}$

S = 1.13 meters.

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