Question

A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no friction between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp? (g = 9.8 m/s2) Group of answer choices

Answer 1

Answer:

$v_f = 3.45 m/s$

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

$mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 10 kg

$\theta = 10^0$

L = 3.5 m

so we will have

$10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)$

$2(9.81) sin10 \times 3.5 = v_f^2$

so final speed is given as

$v_f = 3.45 m/s$