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Setler [38]
3 years ago
10

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

anged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 82.0 J when you compress the springs a distance of 0.220 m from their uncompressed length. What magnitude of force must you apply to hold the platform in this position? How much additional work must you do to move the platform a distance 0.220 m farther? What maximum force must you apply to move the platform a distance 0.220 m farther?
Physics
1 answer:
zmey [24]3 years ago
6 0

Answer:

F =1490.9 N

Explanation:

given,

work done = 82 J

compression length = 0.220 m

for parallel combination

k_{eq}= k_1+k_2

         = 2 k

work done

w = \dfrac{1}{2}k_{eq}x^2

k_{eq} = \dfrac{2W}{x^2}

k_{eq} = \dfrac{2\times 82}{0.22^2}

k_{eq} =3388.43 N/m

force =

F = kx

F = 3388.43 ×0.22

F =745.45 N

the additional work done

final potential energy = \dfrac{1}{2}k_{eq}(2x)^2

                                    = 4 W = 328 J

the additional work = 328 - 82 = 255 J

maximum force

F = k × 2x

F = 2×745.45

F =1490.9 N

                                   

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Answer:

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angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

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Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

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Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

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