Question

A block is projected up a frictionless plane with an initial speed vo. The plane is inclined 30° above the horizontal. What is the approximate acceleration of the block at the instant that it reaches its highest point on the inclined plane?

Answer 1

Answer:

$a=-5 m/s^2$

Explanation:

We are given that Initial speed of block=$v_0$

$\theta=30^{\circ}$

We have to find the approximate acceleration of the block at the instant that it reaches highest point on the inclined plane.

By Newton's second law

$F+mg sin\theta=0$

We know that $F=ma$

$ma=-mg sin\theta$

$a=-gsin\theta$

g=$9.8 m/s^2$

Substitute the values then we get

$a=-9.8 sin30^{\circ}$

$a=-9.8\times \frac{1}{2}=-4.9\approx -5m/s^2$

$sin 30^{\circ}=\frac{1}{2}$

Hence, the approximate acceleration of the block=$-5 m/s^2$

Where negative sign indicates that the block is moving in downward direction.