How Is Newton's Third Law Applied in Calculating Forces on a Lawn Mower?

[F.B]

I hate all of Newton's laws questions. They are so complicated. Can anyone please help me with this question.

A boy pushes a lawn mower (m=17.9 kg) starts from rest across a horizontal lawn by applying a force of 32.9 N straight along the handle, which is inclined at an angle of 35.1 degrees above the horizontal. The magnitude of the mowers acceleration is 1.37 m/s^2, which lasts for 0.58 s, after which the mower moves at a constant velocity. Determine the magnitude of
a)the normal force on the mower
b)the frictional force on the mower
c)the maximum velocity of the answer
d)the force applied by the boy needed to maintain the constant velocity

This is a long question so please help with what you know.

 

[Päällikkö]

Please show what you've done/tried.

Divide the force into horizontal and vertical components. Do remember to take into account the increase in normal force (the push is party directed vertically).

 

[quicknote]

Applying Newton's third law, we know that for every action, there is an equal and opposite reaction. In this scenario, the boy is applying a force of 32.9 N on the lawn mower, but the lawn mower is also applying an equal and opposite force of 32.9 N on the boy. This reaction force is what allows the lawn mower to move forward.

a) The normal force on the mower is the force exerted by the ground on the mower in the opposite direction of the mower's weight. Since the mower is on a horizontal surface and is not accelerating vertically, the normal force will be equal and opposite to the weight of the mower, which is given by mg, where m is the mass of the mower and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the normal force on the mower is 17.9 kg x 9.8 m/s^2 = 175.22 N.

b) The frictional force on the mower is the force that opposes the motion of the mower and is caused by the interaction between the mower and the ground. This force will be equal and opposite to the force applied by the boy on the mower, which is 32.9 N. Therefore, the frictional force on the mower is also 32.9 N.

c) The maximum velocity of the mower can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time. Therefore, v = 0 + (1.37 m/s^2 x 0.58 s) = 0.7966 m/s.

d) Since the mower is now moving at a constant velocity, the net force on the mower is 0. This means that the force applied by the boy on the mower must also be 0. In other words, the force required to maintain the constant velocity is 0 N.