A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m, the coefficient of friction is 0.20.
What is the final speed of the crate after being pulled 21.0m?

18 m/s

Explanation:

Force is given by mass × acceleration

F = ma

a = F/m

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

a = 225/26.0 m/s²

By the equation of motion, v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled.

Using the values for the first part of the motion,

v² = 0² + 2 × 8.65 × 11.0

v = 13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, v² = u² + 2as

v² = 13.8² + 2 × 6.69 × 10 = 324.24

v = 18 m/s

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

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