Two forces, of magnitudes F 1 = 75.0 N and F 2 = 30.0 N , act in opposite direct

Question

Two forces, of magnitudes F1 = 75.0N  and F2 = 30.0N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1) Initially, the center of the block is at position xi = -1.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00cm .

Two forces, of magnitudes F1 = 75.0N and F2 = 30.0N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1) Initially, the center of the block is at position xi = -1.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00cm . Find the work W1 done on the block by the force of magnitude F1 = 75.0N as the block moves from xi = -1.00cm to xf = 3.00cm . Find the work W2 done by the force of magnitude F2 = 30.0N as the block moves from xi = -1.00cm to xf = 3.00cm . What is the net work Wnet done on the block by the two forces? Determine the changeKf?Ki in the kinetic energy of the block as it moves from xi = -1.00cm to xf = 3.00cm .

Explanation / Answer

Given forces, F1 = 75 N , F2= - 30 N (opposite direction) Initially, the center of the block is at position, xi = -1.00 cm At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00 cm Hence,work W1 done on the block by the force of magnitude F1 =75 N as the block moves from xi = -1.00 cm to x f= 3.00 cm W1=F1(xf-xi) =(75 N)[(0.03 m) -(-0.01 m)] = 3 J

Part:B) work W2 done on the block by the force of magnitude F2 = - 30.0 N ( negative sign used for these forces opposite each other) as the block moves from xi = -1.00 cm to x f= 3.00 cm W2=F2(xf-xi) =(-30 N)[(0.01 m) -(-0.03 m)] = - 1.2 J Magnitude, W2=1.2 J

Part:C) Net work done is the sum of these workes, Hence, W=W1+W2 =(3 J)+(-1.2 J) = 1.8 J

Part:D) According to work-energy theorem, Change in kinetic energy is equal to Net work done hence, change (Kf -Ki) in the kinetic energy is, 1.8 J

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