om/courses/1467150/quizzes/1978418 /take Question 2 1 pts A crate resting on a h

Question

om/courses/1467150/quizzes/1978418 /take Question 2 1 pts A crate resting on a horizontal floor uis-0.61, -0.3 ) has a horizontal force F-49 Newtons applied to the right. This applied force is the maximum possible force for which the crate does not begin to slide. If you applied this same force after the crate is already sliding, what would be the resulting acceleration (in meters/second2)? Question 3 1 pts A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.51 radians with the horizontal. The frictional coefficients between the crate and plane are (,-088, k-048), what is the magnitude of the acceleration (in meters/second, of this crate as it slides up the incline? Question 4 1 pts

Explanation / Answer

Given

applied force on cart is F = 49 N

coefficient of static friction is mue_s = 0.61 , mue_k = 0.3

Question 2

force on cart when it was not moving at all by the applied force of F is equal to frictional force

F = mue_s*mg cos theta

49 = mue_s *mg cos theta

49 = mue_s *mg cos theta

49 = 0.61*mg*1

mg = 49/0.61

mg = 80.33 N

m = 80.33/9.8 kg = 8.197 kg

now the force applied the cart is moving then

F = mue_k*mg cos theta

F = ma - mg cos theta  

49 = 8.197*a - 80.33 cos0

a = 15.77 m/s2

Question3

the force acting on the crate is when sliding up is  

F = mg sin theta - mue_k*mg cos theta

a = g sin theta - mue_k*g cos theta

a = 9.8 sin 0.51 + 0.48*9.8 cos 0.51 m/s2

a= 8.8895 m/s2

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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