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

`v_f = 3.45 m/s`

Step-by-step 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 = (1)/(2)mv_f^2 - (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 * 3.5 = (1)/(2)(10)(v_f^2 - 0)`

`2(9.81) sin10 * 3.5 = v_f^2`

so final speed is given as

`v_f = 3.45 m/s`