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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aristocles aristocles · Answer 1 · 2018-10-24T03:21:19+02:00

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

[tex]v_i = 1.5 m/s[/tex]

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

[tex]a = -\frac{F}{m}[/tex]

given that

F = 26 N

m = 22 kg

[tex]a = - \frac{26}{22} = -1.18 m/s^2[/tex]

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

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 1.5^2 = 2*(-1.18)*d[/tex]

[tex]d = 0.95 m[/tex]

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

dsdrajlin dsdrajlin · Answer 2 · 2022-02-25T12:30:49+01:00

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.

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,

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: https://brainly.com/question/1021839