A 89.0 N grocery cart is pushed 13.0 m by a shopper who exerts a constant horizontal force of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is its final speed?
KE gained=workinput
1/2*89*Vf^2=40*13
solve for Vf
thanks so much
To find the final speed of the grocery cart, we can use Newton's second law of motion and the equations of motion.
1. First, let's calculate the acceleration of the cart using Newton's second law: F = ma
The force exerted by the shopper is 40.0 N and the mass of the cart is given by:
mass = force / acceleration due to gravity = 89.0 N / 9.8 m/s^2
2. Once we have the acceleration, we can use the equation of motion to find the final velocity:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown),
u = initial velocity (which is 0 m/s since the cart starts from rest),
a = acceleration (calculated in step 1),
s = displacement (given as 13.0 m).
By rearranging the equation, we have:
v^2 = 2as
3. Substitute the values and calculate:
v^2 = 2 * a * s
v^2 = 2 * (acceleration) * (displacement)
v^2 = 2 * (calculated acceleration) * (13.0 m)
v^2 = (calculated acceleration) * (26 m)
4. Finally, take the square root of both sides to find the final velocity:
v = √[(calculated acceleration) * (26 m)]
Plug in the value for the calculated acceleration in step 1 and solve for v.