Carrie is pushing a 39.8 kg shopping cart at a speed of 0.744 m/s. If she applies a force of 6.59 N, for a period of 2.05 seconds, what will be the speed of the shopping cart after this force is applied?

By applying a force of 6.59-N to a 39.8-kg shopping cart, Carrie accelerates the cart with magnitude a such that

6.59 N = (39.8 kg) a

⇒   a ≈ 0.166 m/s²

so that over a period of 2.05 seconds, the cart's speed increases to

0.744 m/s + a (2.05 s) ≈ 1.08 m/s

Or, if you've been introduced to the concept of momentum and impulse, we get the shopping cart's initial momentum,

p₁ = (39.8 kg) (0.744 m/s) ≈ 29.6 kg•m/s

and its final momentum when it attains speed v,

p₂ = (39.8 kg) v

Then the change in momentum of the cart is

∆p = p₂ - p₁ = (39.8 kg) (v - 0.744 m/s)

and this is equal to the impulse J due to the force Carrie applies,

J = F ∆t = (6.59 N) (2.05 s) ≈ 13.5 kg•m/s

Solve for v :

J = ∆p

⇒   13.5 kg•m/s ≈ (39.8 kg) (v - 0.744 m/s)

⇒   v ≈ 1.08 m/s

80cm/s.

24cm/s.

53.3cm/s.