Work Done by Friction
Trending Questions
- 16.6 m
- 22.3 m
- 28.4 m
- 25.0 m
The work done by kinetic friction on an object is always negative.
True
False
Calculate the work done by kinetic friction in 10 sec (g=9.8 m/s2)
- 246.96 J
- −246.96 J
- 882 J
- −882 J
- 9.82 J
- 4.94 J
- 2.45 J
- 1.96 J
In a children's park, there is a slide which has a total length of 10 m and a height of St m (figure 8-E3). Vertical ladder are provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. (a) the work done by the boy on the ladder as he go: up, (b) the work done by the slide on the boy as he cornea down, (c) the work done by the ladder on the boy as he goes up. Neglect any work done by forces inside the body of the boy.
A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (figure 8-E1). A constant horizontal force F acting on the lower block produces an acceleration F2(m+M) in the system, the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the force of friction on the smaller block by the bigger block during a displacement d of the system.
A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released from rest, form the point P and it comes to rest a point R. The energies lost by the ball, over the parts PQ and QR of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction μ and the distance x (= QR), are respectively close to
0.2 and 3.5 m
0.29 and 3.5 m
0.2 and 6.5 m
0.29 and 6.5 m
Does friction act when an object is at rest?
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/841073/original_7s.png)
- 1250 J
- −1250 J
- 2500 J
- −2500 J
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/841156/original_oo.png)
- 320 J
- 400 J
- 480 J
- 360 J
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/815528/original_9.png)
- 0 J
- 10 J
- −5 J
- 5 J
A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by the friction).
A block of mass 250 g slides down an incline of inclination 37∘ with a uniform speed. Find the work done against the friction as the block slides through 1.0 m.
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/835917/original_5q.png)
- −mg vt sinθ cosθ
- −mg vt cos2θ
- mg vt sinθ cosθ
- −mg vt sinθ
(Take g=10 m/s2)
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/792224/original_7.png)
- 12 N
- 32 N
- 16 N
- 24 N
- 5000 J
- −5000 J
- −1000 J
- 1000 J