Question
I will add more points if you get em right.
A uniform solid sphere rolls without slipping down a 25 degree inclined plane. What is the acceleration of the sphere's center of mass? m/s2 A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. 3.6 J of work is required to compress the spring by 0.2 m. If the mass is released from rest with the spring compressed 0.2 m. it experiences a maximum acceleration of 16 m/s2. Find the value of the mass. kg A 34 g bullet strikes a 0.5 kg block attached to a fixed horizontal spring whose spring constant is 5.6 times 103 N/m and sets it into vibration with an amplitude of 19 cm. What was the speed of the bullet before impact if the two objects move together after impact? m/s Find the coefficient of kinetic friction between a 4.25 kg block and the horizontal surface on which it rests if a 77.0 N/m spring must be stretched by 6.40 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction. A 1.75 kg box rests on a plank that is inclined at an angle of 25 degree above the horizontal. The upper end of the box is attached to a spring with a force constant of 16 N/m, as shown in the figure. If the coefficient of static friction between the box and the plank is 0.85. what is the maximum amount the spring can be compressed and the box remain at rest? m
Explanation / Answer
1)
a = g sin(25) = 9.8*sin25 = 4.14 m/s2
2)
K = 0.5 k x^2
==> k = 2K/x^2 = 2*3.6/(0.2*0.2) = 180 N/m
a_max = A w^2
==> w^2 = a_max/A = 16/0.2 = 80
==> m = k/w^2 = 180/80 = 2.25 Kg
3)
U = 0.5 k A^2 = 0.5*5.6e3 * 0.19*0.19 = 101.08 J
v = sqrt(2K/(m+M)) = sqrt(2*101.08/(0.5+0.034)) = 19.457 m/s
m vi = (M+m) v
==> vi = (M+m)/m v = (0.5+0.034)/0.034 * 19.457 = 306 m/s
4)
k x = u m g
==> u = kx/mg = 77*0.064/()4.25/9.8 = 0.118
5)
m g sin25 + k x = u m g cos25
==> x = (u m g cos25 - m g sin25)/k = (0.85*1.75*9.8*cos25 - 1.75*9.8*sin25)/16 = 0.373 m