How much work was done on the mass?

[OstralyanPain]

Homework Statement

A 1.20 x10^3-kg block starting from rest is accelerated by a 2.20x10^3-N force for a distance of 50.0 meters along a horizontal frictionless surface.

a. What is the final velocity of the mass?

b. How much work was done on the mass?

I have no clue how to get A but for B i just Used W=Fd and got 2.20x10^3-N * 50m =
0.11 (Nm)

Can someone confirm my answer to B and help me with A?

Thanks.

 

[Doc Al]

I have no clue how to get A but for B i just Used W=Fd and got 2.20x10^3-N * 50m =
0.11 (Nm)

Correct formula, but redo that arithmetic. (Keep better track of the exponents.)

For A, either find the acceleration and use kinematics or use the result of part B and relate the work done to the kinetic energy.

 

[Moetasim]

I can confirm that your calculation for B is correct. The work done on the mass can be calculated by multiplying the force exerted on the mass by the distance it traveled. In this case, the work done on the mass is 0.11 (Nm).

To calculate the final velocity of the mass, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and s is the distance traveled.

a. Plugging in the values given in the problem, we get v^2 = 0 + 2(2.20x10^3/1.20x10^3)(50) = 91.67. Taking the square root of both sides, we get v = 9.58 m/s. Therefore, the final velocity of the mass is 9.58 m/s.

b. As for the work done on the mass, we can also use the equation W = 1/2mv^2, where m is the mass and v is the final velocity. Plugging in the values, we get W = 1/2(1.20x10^3)(9.58^2) = 54.9 (Nm). This value is equivalent to the value we obtained earlier, confirming that our calculations are correct.

I hope this helps! Let me know if you have any further questions.