Question #6351

A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal foce of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is the grocery cart's final speed?

Expert's answer

As the acceleration of gravity g = 9.8 m/s^2, then the mass of cart is

m =

98/9,8 = 10 kg.

Using the Newton's Second Law

F = ma. So, the cart's

acceleration is a = F/m = 40/10 = 4 m/s^2.

As the initial speed was zero, we

get the equation for displacement and final speed:

S = a*t^2/2, V = at.

t

= sqrt(2*S/a).

V = a*sqrt(2*S/a) = sqrt(2*S*a) = sqrt(2*12*4) = sqrt(96) =

9.8 m/s.

m =

98/9,8 = 10 kg.

Using the Newton's Second Law

F = ma. So, the cart's

acceleration is a = F/m = 40/10 = 4 m/s^2.

As the initial speed was zero, we

get the equation for displacement and final speed:

S = a*t^2/2, V = at.

t

= sqrt(2*S/a).

V = a*sqrt(2*S/a) = sqrt(2*S*a) = sqrt(2*12*4) = sqrt(96) =

9.8 m/s.

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