Question

A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firmly in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g

Answer 1

Answer:

d) greater than g

Explanation:

You have to apply Newton's Second Law to have an idea about the value of the block's aceleration.

∑F = ma

Where m is the mass, a is the acceleration and F represents all the forces involved.

You have to draw a free body diagram for the block at the top of the loop. (See the attachment)

Applying Newton's Second Law:

y: N+mg=ma

Where N is the normal force (the contact force that the loop's surface applies to the block) and mg is the weight.

Therefore a= $\frac{N}{m}$ + g

So, you can notice that the magnitude of the block's acceleration is greater than the acceleration of gravity because of the effect of the magnitude N/m.

Answer 2

The magnitude of the acceleration of the block is greater than that of the $g$ (gravitational acceleration).

What does Newton's Second Law state?

The force exerted on an object is directly proportional to the acceleration of the object.

$F = ma$

Where

$m$ - mass,

$a$ -  acceleration and

$F$ - all the forces involved.

The forces exerted on the block at the top of the loop.

From Newton's Second Law:

$N+mg = ma$

Where

$N$- normal force (contact force applied on the block by loop's surface)

$mg$ - weight

Thus,

$a=\dfrac Nm + g$

Therefore, the magnitude of the acceleration of the block is greater than that of the $g$ (gravitational acceleration).

Learn more about Newton's Second Law:

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