5.A 60.1 kg skier coasts up a snow-covered hill that makes an angle of 24.0° wit

Question

5.A 60.1 kg skier coasts up a snow-covered hill that makes an angle of 24.0° with the horizontal. The initial speed of the skier is 6.74 m/s. After coasting a distance of 1.72 m up the slope, the speed of the skier is 3.48 m/s. Calculate the work done by the kinetic frictional force that acts on the skis. A) 11.9 J B) -959 J C) -75.7 D) -235 J E) -589

6. A 5.00 kg block is moving at 6.00 m/s along a frictionless horizontal surface toward a spring with a force constant of 500 N/m that is attached to a wall. Find the max distance the spring will be compressed when it stops. A) 0.098 m B) 1.7m C) 0.250 m D) 0.600 m E) 0.420 m

7. The potential energy function of an object is , where A is a positive constant.

i) What is the force?

ii) Is the force acting on this object conservative? A) No, because it is negative. B) Yes, because work can be done by the force. C) No, because the total work in a closed loop is zero. D) Yes, because the force is a vector. E) Yes, because it can be expressed as a potential energy function.

8. A 0.101 g spider hangs from the middle of the first thread of its future web. The thread makes an angle of 5.57° with the horizontal on both sides of the spider. What is the tension in the thread? A) 4.95 x 10-4 N B) 5.10 x 10-3 N C) 4.97 x 10-4 N D) 5.15 x 10-3 N E) 2.56 x 10-3 N

Explanation / Answer

5)(E)-589 J

explantion

Wf = KEf + PEf - KEi

Wf = 1/2 x 60.1 x 3.48^2 + 60.1 x 9.81 x 1.72 x sin24 - 1/2 x 60.1 x 6.74^2 = -588.72 J = -589 J

Hence, (E)-589 J

6)(D)0.6 m

explanation

from conservation of energy

1/2 k x^2 = 1/2 m v^2

x = v sqrt (m/k) = 6 sqrt (5/500) = 0.6 m

Hence, (D)0.6 m

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.