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3 years ago

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A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

s on frictionless rollers. the combined mass of the professor and chair is 85.0 kg. she is pushed 2.50 m along the incline by a group of students who together exert a constant horizontal force of 600 n. the professor’s speed at the bottom of the ramp is 2.00 m>s. use the work–energy theorem to find her speed at the top of the ramp.

1 answer:

11111nata11111 [884]3 years ago

7 0

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$

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