Consider a frictionless track ABC as shown in Figure P8.23. A block of mass m1 = 8.00 kg is released from A. It makes a head-on elastic collision at B with a block having a mass of m2 = 14.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision...
A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?
F\DeltaT = \Deltap
p =...
A system consists of three particles, each of mass 4.00 g, located at the corners of an equilateral triangle with sides of 36.0 cm.
(a) Calculate the potential energy of the system.
m = .004, d = .2078
UG = Gm1m2/R
U_total = sum of U_i
U_total = Gm1m2/R + Gm1m3/R + Gm2m3/R
U_total =...
800 kg roller coaster is initially at the top of a rise, at point A. It then moves 125 ft (38.1 m), at an angle of 40.0° below the horizontal, to a lower point, B.
(a) Choose point B to be the zero level for gravitational potential energy. Find the potential energy of the roller coaster-Earth...
(delta KE) = (m2)(g)(h) - (m1)(g)(h)(sin(theta)) - (m)(g)(cos(theta))(uk)(x)
I added in the potential energy for the second block. Still incorrect. I don't understand why I would need tension, nor would I know how to incorporate it into the formula.
Edit: Isn't (m2)(g)(h) the string tension?
A toy cannon uses a spring to project a 5.30 g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 9.00 N/m. When it is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and a constant frictional force of 0.0320 N exists between...
A m1 = 45.0 kg block and a m2 = 105.0 kg block are connected by a string as in Figure P7.44. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 45.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 45.0 kg block as...
A 6.00 kg block is set into motion up an inclined plane with an initial speed of v0 = 8.50 m/s (Fig. P7.23). The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal.
(a) Determine the friction force exerted on the block (assumed...
(m2g)(y) - (uk)(m1g)(x) = .5m2vf2 - .5m2vi2
(5)(9.8)(1.2) - (.2)(3)(9.8)(1.2) = (.5)(5)(vf2) - 0
vf = 4.54
Final answer is incorrect. Any ideas?
Edit: Got it. As you just mentioned, I should have used mass of the system instead of just m2 on the right side.
vf = 3.59
Thanks.
The coefficient of friction (uk) between the 3.00 kg block (m1) and the surface in Figure P7.21 is 0.200. The system starts from rest. What is the speed of the 5.00 kg ball (m2) when it has fallen 1.20 m (y)?
Wnet = \DeltaKE
Um2 - Wfk = .5m2vf2 - .5m2vi2
(m2g)(y) - (uk)(N)(x) = 5m2vf2 -...
T = ma
T = (mass of the system)(acceleration of the system)
m2g matches the direction of the movement, therefore it is positive
fk opposes the movement, therefore it is negative.
T + m2g - fk = (m1 + m2)a