n the diagram above, the net force acting on the 20-kg lawn mower is 51 N and the force of friction opposing the motion is 24 N. a) What is the magnitude of the force the person exerts on the mower?b) Suppose the mower is moving at 1.5 m/s when the person stops pushing. How far will the mower go before stopping?A portion of this content was adapted from College Physics from Open Stax (Links to an external site.). Download for free at https://openstax.org/details/books/College-Physics (Links to an external site.).Group of answer choicesa) 75 Nb) 1.25 ma) 75 Nb) 0.94 ma) 27 Nb) 1.25 ma) 27 Nb) 0.94 m

a) The magnitude of the force the person exerts on the mower can be calculated by adding the net force and the force of friction. The net force is given as 51 N and the force of friction is 24 N.

So, Force exerted = Net force + Force of friction = 51 N + 24 N = 75 N

b) To find out how far the mower will go before stopping, we first need to find the acceleration of the mower. We can use the formula F = ma, where F is the net force, m is the mass and a is the acceleration.

So, Acceleration = Net force / Mass = 51 N / 20 kg = 2.55 m/s²

The force of friction acts in the opposite direction to the motion of the mower, so it decelerates the mower. The deceleration can be calculated using the formula F = ma, where F is the force of friction and m is the mass.

So, Deceleration = Force of friction / Mass = 24 N / 20 kg = 1.2 m/s²

We can use the formula v² = u² + 2as to find the distance the mower travels before it stops, where v is the final velocity (0 m/s, as the mower stops), u is the initial velocity (1.5 m/s), a is the deceleration (1.2 m/s²) and s is the distance.

So, 0 = (1.5 m/s)² + 2 * -1.2 m/s² * s

Solving for s, we get s = (1.5 m/s)² / (2 * 1.2 m/s²) = 0.94 m

So, the answers are a) 75 N and b) 0.94 m.