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3 years ago

9

Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

° below the horizontal. The sled then moves 15.4 m forward. How much work did the skater do on the sled? Assume friction is negligible.
Knowns Unknown

F = ____________ d = _____________ W = ?

θ = ____________

Solve for the Unknown

a. Write the equation for work
b. Insert known quantities

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

Expression of work done is

$W = Fd cos\theta$

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

$W = Fd cos\theta$

here we know that

F = 12.6 N

d = 15.4 m

$\theta = 15.3 degree$

so we will have

$W = 12.6 \times 15.4 cos15.3$

$W = 187.2 J$

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

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$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

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Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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