1. A 3.00 kg block moves down a 25 degree incline at constantvelocity. Find the

Question

1. A 3.00 kg block moves down a 25 degree incline at constantvelocity. Find the coefficient of kinetic friction between theblock and the incline. (g= 9.81 m/s2)

2. A crate of books that weighs 200 N is dragged horizontallyacross the hall floor. The crate is pulled with a force of 125 N atan angle 37 degrees above the horizontal. If the crate is initiallyat rest and the coefficient of friction is .085, how long does ittake to drag the crate of books 15.7 m?

3. A 185 kg block slides down a 12 m incline of 55 degrees abovethe surface. The coefficient of kinetic friction is .345. Find thevelocity of the block when it reaches the bottom.

4. An 8.5 kg block compresses a spring with k= 555N/m a distanceof .735 m. The coefficient of kinetic friction between thehorizontal surface and the block is .456. How far does the blockmove before it comes to a stop?

5. A bullet with a mass 4.5 g is fired at 535 m/s horizontallyinto a 2 kg block attached to a horizontal spring that is connectedto a wall. The spring constant is 400 N/m. What is the maximumcompression of the spring?

6. At a Volvo automobile plant, a crash test was performed. AnSUV of 1800 kg traveling at -25 m/s collides with a wall andrebounds. If the average force is 341,380 N, and the impulse of theSUV on the wall is 49,500 N,

      a) What was the SUV'srebounding velocity?

      b) What was the time ofimpact?

Explanation / Answer

a. mgsin=kmgcos k=tan=tan25=0.466 b. The crate moves with the resultant acceleration as, 125cos37-0.085*200=ma m=200/9.8kg From this we get a. The velocity for 17.5m is v2-u2=2as The time taken is v=u+at c. The work done=change in KE W=KE mgsin55*12=1/2*mv2 From this we get v. d. The PE of spring=frictional work done due to block 1/2*kx2=k*mg*d 1/2*555*0.7352=0.456*8.5*9.8*d e. Apply the law of conservation of momentum then equate its KEwith springs PE. f. Impulse I=F*t From this we get time taken Then I=change in mometum I=mv2-mv1 From this we get the rebounding velocity. Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework." From this we get v. d. The PE of spring=frictional work done due to block 1/2*kx2=k*mg*d 1/2*555*0.7352=0.456*8.5*9.8*d e. Apply the law of conservation of momentum then equate its KEwith springs PE. f. Impulse I=F*t From this we get time taken Then I=change in mometum I=mv2-mv1 From this we get the rebounding velocity. Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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