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Leni [432]
3 years ago
14

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.

Physics
1 answer:
SpyIntel [72]3 years ago
7 0

Answer:

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

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.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → W=2,55.0,71=1,8105\frac{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 = \frac{1}{2} mV^{2} + mgh

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

In a triangle \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

In our situation

\frac{0,71}{sin90} =\frac{h}{sin32} → h=0,376

Therefore

1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376

→ V=2,416

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