a student pushes a 40 kg lawnmower from rest across a rough surface with a force of 200 N at an angle of 60 degrees below the horizontal. the lawnmower is moved 3 meters. the coefficient of kinetic friction is 0.3.

how much work did the student do on the lawn mower moving it forward 3 meters?

how much work was transfered from the mower to the ground through the friction force?

what is the speed of the lawnmower at the end of the 3 m of motion?​

Well, let's crank up the comedy engine and mow down these questions!

To find out how much work the student did on the lawnmower, we need to determine how much of the applied force was in the direction of motion. Since the force is at an angle of 60 degrees below the horizontal, only a portion of it will be used to move the lawnmower forward. We can calculate that using the formula:

Work = Force * Distance * cos(angle)

Work = 200 N * 3 m * cos(60)

Work = 200 N * 3 m * 0.5

Work = 300 J

So, the student did 300 J of work on the lawnmower. That's a real work of art!

Now, let's move on to the work transferred from the mower to the ground through the friction force. The work done by friction can be determined using the equation:

Work = Force of Friction * Distance

The force of friction can be calculated by multiplying the coefficient of kinetic friction (0.3) with the normal force. Since the lawnmower is on a rough surface, the normal force is equal to the weight of the lawnmower.

Force of Friction = μ * Normal force

Normal force = Mass * Gravity

Normal force = 40 kg * 9.8 m/s²

Force of Friction = 0.3 * (40 kg * 9.8 m/s²)

Work = (0.3 * 40 kg * 9.8 m/s²) * 3 m

Work = 352.8 J

So, the work transferred from the lawnmower to the ground via friction is a total of 352.8 J. That's a slippery slope!

Finally, let's determine the speed of the lawnmower at the end of the 3 m of motion. We can use the work-energy principle, which states:

Work = ΔKinetic Energy

The work done by the student on the lawnmower equals the change in kinetic energy of the lawnmower. So, we can set up an equation:

300 J = (1/2) * Mass * Velocity²

300 J = (1/2) * 40 kg * Velocity²

Velocity² = (300 J * 2) / (40 kg)

Velocity = sqrt[(300 J * 2) / (40 kg)]

Velocity ≈ 6.71 m/s

So, the lawnmower has a speed of approximately 6.71 m/s at the end of the 3 m of motion. That's mowtastic!

I hope this green and clean answer brings a smile to your face. Keep mowing forward, my friend!