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2 years ago

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A shopper at the Fries’ market place pushes a 13.0-kg shopping cart at a constant velocity for a distance of 58.7 m on a level s

urface. He pushes in a direction 24.50 below the horizontal. A 53.8 N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts on the cart? (b) What is the work done by the pushing force? (c) What is the work done by the frictional force? (d) What is the work done by the gravitational force?

1 answer:

Gnom [1K]2 years ago

4 0

We do a force summatory, then,

a) $\sum F=0$

$Fcos(24.5)-F_f= 0$

$Fcos(24.5)-53.8= 0$

$F= \frac{53.8}{cos(24.5)}$

$F=59.12 N$

(b) For the Work we need the equiation with the variable in X.

$W= F (d cos(24.5))$

$W= 59.12 (58.7) cos 24.5$

$W=3158.1 J$

(c) However the Frictional Force Work,

$W= F_f d$

$W= -53.8(58.7)$

$W=-3158.1 J$

(d) Now the work by gravity

$W_g= F d (cos(90))=0$

$W_g=0$

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$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$

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c)

No, The event are not simultaneous i.e they did not occur at the same time.

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