a) 44.0 b) 54.0 c)64.8 d) 52.8 4. A car with mass 953 kg passes over a bump in a
ID: 1783683 • Letter: A
Question
a) 44.0 b) 54.0 c)64.8 d) 52.8 4. A car with mass 953 kg passes over a bump in a road that follows the arc of a circle of radius 59.5 m as shown in the figure. Find the normal force (in kN) on the car by the road when the car is at the top of the bump as shown below if the speed of the car is 15.0 m/s 953 kg a) 5.74 b) 10.46 c) 12.94 d) 3.87 5. A 15.0-kg object is pulled along a horizontal surface at a constant speed by a 70-N force acting 30 above the horizontal. How much work (in Nam) is done by this force as the object moves 4.0 m in the horizontal direction? 15 kg a) 242.5 b) 85.5 e) 68.3 d) 303.1 6. A net force F=(6-2 ) N acts on a 40-kg object as it moves from the ongin to the point s=(Si + 4j) m. What is the work done by this force? a) 36 b) 10 c) 22 d) 48 7. A 2.0-kg particle has an initial velocity of (3i - 4j) m/s. Some time later, its velocity is (8i+6) m/s. What is the total work (in J) was done on this particle during this time interval? a) 53 b) 80 c) 29 d) 75 8. A parachutist of mass 49.3 kg jumps out of an airplane at a height of 842 m and lands on the ground with a speed of 7.68 m/s. How much energy (in kJ) was lost to air friction during this jump a) 405.35 b) 296.32 c) 728.12 d) 593.89Explanation / Answer
4. in vertical,
Fnet = m g - N = m a_c = m v^2 / r
(953 x 9.81) - N = 953 (15^2) / 59.5
N = 5745 N Or 5.74 kN
Ans(A)
5. W = F.d = F d cos(theta)
W = 70 x 4 x cos30 = 242.5 J Ans(A )
6. W = F.s = (30) - 8 = 22 J Ans(c)
7. work done = m (vf^2 - vi^2) /2
= 75 J
Ans(d)
8. (49.3 x 842 x 9.8) - W = (49.3 x 7.68^2 / 2)
W = 405.35 x 10^3 J = 405 .35 kJ
Ans(a)
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