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A horizontal force of 200N is applied to move a 55-kg cart (initially at rest) across a...
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You need to use the equation of kinetic energy, such that:
`KE = (1/2)mv_f^2`
You need to evaluate the final velocity of the car using the following equation, such that:
`v_f^2 = v_i^2 + 2a*d`
`v_i` represents the initial velocity
a represents the acceleration s represents the distance
Since the cart is initially at rest, `v_i = 0` , such that:
`v_f^2 = 2*a*10 `
You may evaluate the acceleration using the Newton's second law, such that:
`F = m*a => a = F/m => a = 200/55 => a = 3.63 m/s^2`
`v_f^2 = 2*3.63*10 => v_f^2 = 72.6 (m^2/s^2)`
`KE = (1/2)*55*72.6 => KE = 1996.5 J`
Hence, evaluating the kinetic energy of the cart, yields `KE = 1996.5 J.`
Posted by sciencesolve on October 26, 2013 at 5:49 PM (Answer #1)
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