1111_Fa16_Exam2_B_P3_solns.pdf

# Next we use the m 2 equation to solve for f t m 2 g f

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Next we use the m 2 equation to solve for F T : m 2 g - F T = m 2 a = F T = m 2 g - m 2 a. Combining this with our last result, we can solve for a : m 2 g - m 2 a - μ k m 1 g = m 1 a = m 2 a + m 1 a = m 2 g - μ k m 1 g = a = m 2 - μ k m 1 m 2 + m 1 g. Plugging in our numbers: a = 2 . 20 kg - (0 . 190)(6 . 40 kg) 2 . 20 kg + 6 . 40 kg (9 . 81 m/s 2 ) 1 . 12 m/s 2 . Copyright c 2016 University of Georgia. 6

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Physics 1111 Exam #2B-11:00 Solutions 13 October 2016 Last Name: IV: Out of Gas (21 points) Tom and Ray push their 760 . 0 kg car from rest along a flat road towards the nearby gas station. Both push horizontally, and Tom pushes with a force of 150 N at an angle of 12 . 0 from the axis of the car. Ray pushes with a force of 170 N at an angle of 14 . 0 from the axis of the car. The gas station is 330 m away. There is a friction force of 305 N. (a) How much work do Tom and Ray do on the car over this distance? (7 pts) The work done by a constant force is given by W = F Δ r cos θ , so we just need to apply this twice. W T = F T d cos θ T = (150 N)(330 m)(cos 12 . 0 ) = 48418 . 3 J 4 . 55 × 10 4 J W R = F R d cos θ R = (170 N)(330 m)(cos 14 . 0 ) = 54433 . 6 J 5 . 11 × 10 4 J Adding these results together gives W TR = W T + W R = 48418 . 3 J + 54433 . 6 J 1 . 03 × 10 5 J . (b) What is the total work done on the car at the end of this distance? (7 pts) Friction acts opposite to the direction of motion in this case, so it does negative work on the car. W f = F f d cos 180 = (305 N)(310 m)( - 1) = - 100650 J . To find the total work done on the car, we add up the work done by all the individual forces: W tot = W T + W R + W f = 48418 . 3 J+54433 . 6 J - 100650 J = 2201 . 9 J 2 . 20 × 10 3 J . (c) What is the speed of the car at the end of this distance? (7 pts) This can be done with Newton’s Second Law and kinematics, but it’s easier to use the Work-Energy Theorem: W tot = Δ K = 1 2 mv 2 f - 1 2 m (0) 2 = v f = r 2 W tot m Plugging in numbers, we have v f = s 2(2201 . 9 J) 760 kg 2 . 41 m/s . Copyright c 2016 University of Georgia. 7
Physics 1111 Exam #2B-11:00 Solutions 13 October 2016 Last Name: V: Roundabout (16 points) The driver of a 13 , 500 kg bus wants to make a turn of radius 28 . 0 m on a flat road so that the passengers feel a horizontal acceleration of no more than 0 . 360 g , where g is the magnitude of the acceleration due to gravity. (a) What is the maximum speed that the driver can maintain around the curve? (7 pts) The bus is making a circular turn, which means there must be a radial acceleration. this acceleration must not exceed the stated value, so we can set it equal to the maximum and solve for the speed: a r = v 2 max R = 0 . 360 g v max = p 0 . 360 gR = q (0 . 360)(9 . 81 m/s 2 )(28 . 0 m) 9 . 94 m/s . (b) What is the minimum value for the coefficient of friction between the tires and the road needed to make the turn at this speed? mg F N f s +x +y a f s a v (9 pts) The static friction force on the tires from the road is the force that causes the radial acceleration and allows the bus to turn. Since the acceleration is along the radial direction, let’s set that as one of the coordinate axes. The other forces acting on the bus are vertical, so our second axis will be vertical. We can then set up two Newton’s Second
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