A shopper in a supermarket pushes a loaded 31 kg cart with a horizontal force of 15 N. What is the distance moved the cart in the following cases:
a) Disregarding friction how far will the cart move in 4.3 s, starting from rest?
b) How far will the cart move in the 4.3 s if the shopper places a(n) 88 N child in the cart before pushing it?
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The shopper pushes the cart with a force of 15 N. The cart is loaded to make its mass equal to 31 kg. The cart starts from rest.
The force applied results in an acceleration given by a = force/mass. The distance moved by an object starting from rest and with an acceleration a in t seconds is given by (1/2)*a*t^2
For a force of 15 N, the acceleration is a = 15/31
In 4.3 s the distance moved D = (1/2)*(15/31)*(4.3)^2
=> 4.47 m
The mass of a child weighing 88 N is 88/9.8. If the child is placed in the cart the mass of the cart becomes 31 + 88/9.8 kg.
The distance moved by the cart in 4.3 seconds in this case is (1/2)*(15/(31+88/9.8))*4.3^2
=> 3.46 m
In 4.3 seconds the cart moves 4.47 m when it is loaded and the child is not placed in it. And it moves 3.46 m when the child is placed in it.
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