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What is the final speed

Physics
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A(n) 84.7 N grocery cart is pushed 13.4 m along an aisle by a shopper who exerts a constant horizontal force of 44.5 N. The acceleration of gravity is 9.8 m/s 2 . If all frictional forces are neglected and the cart starts from rest. Answer in m/s

Jan 27th, 2015

Please note that if the mass m of your cart is m=84.7 Kg, then, we can say that the magnitude of the acceleration a, of your cart, is fiven by the subsequent formula:

a= F/m, where F is the mafnitude of the force acting on you cart, namely F=44.5 N.

So we have:

a=44.5/87.4= 0.5 m/sec^2.

Now the relationship among the distance d traveled by our cart, the magnitude of the acceleration a of our cart, and the time needed to travel that distance, is:

d=(1/2) a t^2,

from which I can write:

t= sqrt[(2d)/a]

substituting your numerical dat into that formula, we get:

t= sqrt [(2*13.4)/0.5]= 7.32 sec, and the speed v reached by our cart in that time interval (0, t), is:

v= a*t=0.5*7.31=3.66 m/sec.

Jan 27th, 2015

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