Question

Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 m/s. A girl pushes a 21 kg lawn mower as shown in the figure


a)If F = 32 N and θ = 45 ∘ what is the acceleration of the mower? Ignore friction.


b)What is the normal force exerted on the mower by the lawn?

Answer 1

Newton's second law allows us to find the result for the acceleration of the mower is:

         a = 0.8 m / s²

Newton's second law says that the net force is proportional to the mass and the acceleration of the bodies.

             ∑ F = m a

 

where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies.

A free-body diagram is a schematic where the forces are drawn if the details of the bodies, see attached.

Let's write Newton's equation.

x-axis

         Fₓ = ma

y-axis

         N - F_y - w = 0

Let's use trigonometry to find the components of the force.

         Cos 45 = $\frac{F_x}{F}$  

         sin 45 = $\frac{F_y}{F}$  

         Fₓ = F cos 45

         F_y = F sin 45

Let's substitute.

         F cos 45 = m a

         a = $\frac{F \ cos45}{m}$

Let's calculate.  

         a = $\frac{32 \ cos 45}{21}$  

         a = 0.8 m / s²

In conclusion using Newton's second law we can find the result for the acceleration of the mower is:

         a = 0.8 m / s²

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Answer 2

The acceleration of the mower is 1.1m/s^2. The normal force exerted on the mower by the lawn is 145.5 N.

a) The force that moves the lawn mower in the horizontal direction is Fcosθ.

Where;

F = 32 N

θ = 45∘

Effective force that moves the mower forward = 32 N cos 45∘ = 22.6 N

Now;

F = ma

m = mass of the lawn mower

a = acceleration

a = F/m

a =  22.6 N/21 kg

a = 1.1 m/s^2

b) The normal force = mgcosθ

FN = 21 kg × 9.80 m/s × cos 45 ∘

= 145.5 N

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