Two forces, of magnitudes F1 = 100 N and F2 = 20.0 N, act in opposite directions

Question

Two forces, of magnitudes F1 = 100 N and F2 = 20.0 N, act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position xi = -3.00 cm. At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00 cm.

 

Two forces, of magnitudes F1 = 100 N and F2 = 20.0 N, act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position xi = - 3.00 cm. At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00 cm. 1. Find the work W_1 done on the block by the force of magnitude F_1 = 100 N as the block moves from x_i = -3.00 cm to x_f = 3.00 cm. 2. Find the work W_2 done by the force of magnitude F_2 = 20.0 N as the block moves from x_i = -3.00 cm to x_f = 3.00 cm. 3. Determine the changeK_{f} - K_{i} in the kinetic energy of the block as it moves from x_i = -3.00 cm to x_f = 3.00 cm.

Explanation / Answer

1) W1 =F1.(xf-xi) = 100 *(.03-(-.03)) = 6 Nm 2) W2 = F2.(xf-xi) = -20. (.03-(-.03)) = -1.2 Nm 3) Kf-Ki = W1+W2 = 6-1.2 = 4.8 N

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