A non-constant force on a particle of mass m is given by F = F,| -+1 where Fo and xo are constants. The knowns are Fo, xo, and m. (a) Find the work done by F in moving the particle from x=0 to x= 2xo- (b) If the particle has mass m and started from rest, what is its final speed?
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- You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m>s. In stopping the ball, your hands and the volleyball descend together a distance of 0.150 m. (a) How much work do your hands do on the volleyball in the process of stopping it? (b) What is the magnitude of the force (assumed constant) that your hands exert on the volleyball?(a) A particle of mass 100 g has an acceleration along the 2-axis given by a = 10 - t. Find expressions for the velocity, v, and the location, r as functions of time (take v = 10 m s-¹ and = 0 at t = 0). (b) What is the furthest (positive) distance from the starting point that the particle attains? (e) What is the work done on the particle between the starting point and the point of furthest distance? (d) What is the power being exerted on the particle at the point (t > 0) when it again passes through x = 0?A force F→ in the positive direction of an x axis acts on an object moving along that axis. If the magnitude of the force is F = 7.2e-x/2.5 N, with x in meters, find the work done by F→ as the object moves from x = 0 to x = 3.0.
- A 0.200-kg particle has a speed of 2.30 m/s at point O and kinetic energy of 7.10 J at point ®. (a) What is its kinetic energy at O? (b) What is its speed at ®? m/s (c) What is the net work done on the particle by external forces as it moves from toA skateboarder is at the top of a frictionless ramp defined by the function y = z². The top of the ramp is located at (-5m, 25m), and the skateboarder skates to the bottom of the ramp, (0m, 0m). (a) What nonconservative forces are in this problem? (b) What is the total nonconservative work done? (c) Is mechanical energy conserved? (d) What is the skateboarder's speed at the bottom of the ramp?1) a) What is the work done when a box weighing 10N is lifted up onto a shelf 2m above the floor? b) If we consider the acceleration due to gravity to be 10 ms^-2 and the same box (weighing 10N) moves at 10 ms^-1 how much kinetic energy is used? c) How much potential energy does the box have when it is on the shelf? d) Thinking about the conservation of energy, if a block of mass 20kg falls off a shelf that is 3m above the ground what is the velocity of the box just before it hits the ground?
- A cart (mass m = 0.5 kg) moves along the x-axis. It is accelerated by a net force F, which varies with position: (1) What is the net work done on the car over the 5-meter displacement? (2) If the car started from rest at x = 0, what is its speed at x = 5 m?A car with mass 1520 kg is traveling down the highway at a speed of 18 m/s when the driver slams on the brakes due to an accident up ahead. The car eventually comes to rest.According to the work-energy theorem the work is related to the change in kinetic energy, Wnet = Δ KE = KEfinal - KEinitial.(a) Using the work-energy relationship, determine how much net work is done on the car from the brakes? Report the magnitude of the the work (positive value) even though the work from brakes will be negative since the car is slowing down._____ J(b) The brakes apply a force of 24000 N to the car in order to make it stop. Using the fact that W = F d and the fact that the you found the work done (magnitude) by the brakes in part (b), determine the stopping distance, d, of the car.____ mSuppose a box of mass 5 kg is initially at rest on a frictionless horizontal surface. A constant force of 10 N is applied horizontally to the box for 4 seconds, and then a 7 N force of the same direction is applied for another 4 seconds. (a) Find the total distance that the box traveled. (b) Find the total work done on the box.
- A mountain climber is about to haul up a 90-m length of hanging rope. How much work will it take if the rope weighs 0.65 N/m? Set up the integral that gives the work done. (Type an exact answer, using radicals as needed.)(a) A force F = (2xî + 4yj), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x-direction from the origin to x = 4.63 m. Find the work W = F. dr done by the force on the object (in J). (b) What If? Find the work W = F• dr done by the force on the object (in J) if it moves from the origin to (4.63 m, 4.63 m) along a straight-line path making an angle of 45.0° with the positive x-axis.The question is attatched down below.