# A horizontal force of 200N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final kinetic energy of the cart?

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### 2 Answers

m= 55

F= 200

a= F/m = 3.63

s= 10m

`v^2=u^2 +2as`

` ` = 0 + 2* 3.63 * 10

= 72.6

`kappa=1/2 mv^2`

= 55 * 72.6 * 1/2

=* 1996.5 J*

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.`