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

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: link

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

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

given that

F = 26 N

m = 22 kg

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

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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14