Question

A 200-N sled of slides down a frictionless hillside that rises at 37° above the horizontal. What is the magnitude of the force that the hill exerts on the sled parallel to the surface of the hill?

Answer 1

Answer:

120 N

Explanation:

Given the angle of inclination of 200 N sled is 37°.

We need to find the magnitude of the force that the hill exerts on the sled parallel to the surface of the hill.

We can see from the diagram, the force that the hill exerts on the sled parallel to the surface of the hill is

$mg\times sin(37 \°)$

As we have given the weight of the sled was 200 N. So, $mg=200\ N$

Now,

$200\times sin(37 \°)\\=200\times 0.601\\=120\ N$

So, 120 N is the force that sled exerts parallel to the surface of the heal.

Answer 2

The magnitude of force will be "159.727 N".

Given:

• Weight of sled, mg = 200 N
• Angle, $\Theta$ = 37°

Now,

The magnitude of force that hill exerts parallel to the surface will be:

= $mg Cos \Theta$

By substituting the values, we get

= $200 Cos 37^{\circ}$

= $159.727 \ N$

Thus the above answer is right.

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