Physics, 06.07.2019 05:30 SassyChumpkins

Suppose the mower is moving at 1.5 m/s when the force f is removed. how far (in m) will the mower go before stopping? the frictional force is a constant 26 n and mass of mower is 22kg

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Answer from: franklynvaldez01

A 22-kg mower, initially moving at 1.5 m/s, will displace for 0.94 m, when a frictional force of 26 N is applied.

What is a uniformly accelerated motion?

A uniformly accelerated motion is the one in which the acceleration of the particle throughout the motion is uniform.

A mower is moving at 1.5 m/s and the force F is removed. It will experience a deceleration (a) due to the frictional force (f).

Given its mass (m) is 22 kg, we can calculate the deceleration using Newton's second law of motion.

a = f / m = 26 N / 22 kg = 1.2 m/s²

Then, we can calculate the displacement after the force F is removed, using the following SUVAT equation.

v² = u² - 2 × a × s

s = u² - v² / 2 × a

s = (1.5 m/s)² - (0 m/s)² / 2 × (1.2 m/s²)

s = 0.94 m

where,

v is the final speed.u is the initial speed.a is the deceleration.s is the displacement.

A 22-kg mower, initially moving at 1.5 m/s, will displace for 0.94 m, when a frictional force of 26 N is applied.

Learn more about friction here: link

Answer from: rachael382

initial speed of the lawn mover when external force is removed from it is given as

v_i = 1.5 m/s

now at this time only friction force is acting on it in the opposite direction to its velocity

this force will cause acceleration in opposite direction which will reduce the speed

$a = -\frac{F}{m}$

given that

F = 26 N

m = 22 kg

$a = - \frac{26}{22} = -1.18 m/s^2$

now we can use kinematics to find the distance moved by the lawn mover

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2*(-1.18)*d

d = 0.95 m

so it will move a distance d = 0.95 m before stop

Answer from: Quest

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answer: 0 meters/second^2

Answer from: Quest

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Suppose the mower is moving at 1.5 m/s when the force f is removed. how far (in m) will the mower go...

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