Problem 3P

If you tried to smuggle gold bricks by filling your backpack, whose dimensions are 60 cm × 28 cm × 18 cm, what would its mass be?

ANSWER: STEP 1:- For the above problem, the acceleration due to gravity value has taken as g = 10 m/s .2 The downward force which is 2500 N. The displacement is 5 meters. Work done by the force is, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. 0 The angle between force and displacement here is 0 , as both are in the same direction. Then work done will be, W = 2500 N × 5 m × cos 0 = 2500 N × 5 m × 1 = 12500 jules. STEP 2:- 0 The tension T m1kes an angle 60 the horizontal, so the angle with the displacement which is vertically downwards will be, 0 0 0 60 + 90 = 150 . The magnitude of force T is11830 N. So, the work done by the force T wi1l be, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. The angle between force and displacement here is 150 . 0 Then work done will be, W = 1830 N × 5 m × cos 150 = 1830 N × 5 m × ( 0.866). = 7924.13 jules The minus sign is trivial, as the cosine of 150 is negative. So the work is negative. STEP 3:- 0 The tension T m2kes an angle 45 the horizontal, so the angle with the displacement which is vertically downwards will be, 0 0 0 45 + 90 = 135 . The magnitude of force T is 1295 N. So, the work done by the force T wi2l be, W = F. ds = F. S = |F||S|cos As the force is independent of the displacement here or constant, we took it out of the integral sign and the integral over the infinitesimal displacement vector will become the total displacement. 0 The angle between force and displacement here is 150 . Then work done will be, 0 W = 1295 N × 5 m × cos 135 = 1295 N × 5 m × ( 0.707). = 4578.51 jules The minus sign is trivial, as the cosine of 135 is negative. So the work is negative. CONCLUSION: The work done by the 3 forces we got as, 12500 jules, 7924.13 jules and 4578.51 jules.