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amid [387]
2 years ago
7

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

rce on the shopping cart ? b) What is the force of friction on the shopping cart ? c) What will happen to the shopping cart if you come across a wet portion of the parking lot and you continue to apply the same force ?
Physics
1 answer:
AlekseyPX2 years ago
4 0

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

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Explanation:

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Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

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\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

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\implies\Delta x=50.4\,\mathrm m

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