Question
i need some help
A crate of mass 11.0 kg is pulled up a rough incline with an initial speed of 1.60 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 20.6 degree with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.98 How much work is done by the gravitational force on the crate? Determine the increase in internal energy of the crate-incline system owing to friction. How much work is done by the 102-N force on the crate? What is the change in kinetic energy of the crate? What is the speed of the crate after being pulled 4.98 m? A 5.80-kg particle moves along the x axis. Its position varies with time according to x = t + 5 t3, where x is in meters and t is in seconds. Find the kinetic energy of the particle at any time t. (Use the following as necessary: t.) Find the magnitude of the acceleration of the particle and the force acting on it at time t. (Use the following as necessary: t.) Find the power being delivered to the particle at time t. (Use the following as necessary: t.) Find the work done on the particle in the interval t = 0 to t = 3.30 s. After a 0.270-kg rubber ball is dropped from a height of 1.95 m, it bounces off a concrete floor and rebounds to a height of 1.20 m. Determine the magnitude and direction of the impulse delivered to the ball by the floor, magnitude kg.m/s Estimate the time the ball is in contact with the floor to be 0.04 seconds. Calculate the average force the floor exerts on the ball, magnitude N A neutron in a nuclear reactor makes an elastic, head-on collision with the nucleus of a carbon atom initially at rest. What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (The mass of the carbon nucleus is about 12.0 times the mass of the neutron.) The initial kinetic energy of the neutron is 2.10 x 10'13 J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. A billiard ball moving at 5.80 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.92 m/s at an angle of 32.0 degree with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision, magnitude m/s direction degree (with respect to the original line of motion)
Explanation / Answer
1)
a) W = F d cos theta, where the is the angle between force and direciton of movement
W = 11*9.81*4.98*cos(110.6)=-189 J
b) W friction = u m g d cos theta = 0.4*11*9.81*4.98*cos(20.6)=201.21 J
c) W = 102*4.98=507.96
d) dKE = W net = -189-201.21+507.96=117.75
e) 0.5*11*(v^2-1.6^2)=117.75
v=4.896 m/s
2) a) v = dx/dt = 1 + 15 t^2
so K = 1/2 mv^2 = 0.5*5.8(1+15t^2)^2
b) a = dv/dt = 30 t
F = ma = 5.8*30t
c) P = Fv = 5.8*30*t*(1+15t^2)
d) W = integral of P dt from 0 to 3.3 = integral of 5.8*30*t*(1+15t^2) from 0 to 3.3=78328.8 J
