Question

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55 N and is parallel to the ramp.What is the speed of the block after it travels a distance d = 0.71 m if the block starts from rest? Use work and energy.

Answer 1

The answer is 2,416 m/s. Let's jump in.

Step-by-step explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point. At the rest, block doesn't have kinetic energy and since it is on the reference point(as we decided) it also has no potential energy.

Under the force block gains;

W = F.x →
`W=2,55.0,71=1,8105(N)/(m)`

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE =
`(1)/(2) mV^(2) + mgh`

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle
`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)`

In our situation

`(0,71)/(sin90) =(h)/(sin32)` →
`h=0,376`

Therefore

`1,8105=(1)/(2) 0,274V^(2) +0,274.9,81.0,376`

→
`V=2,416`