A single force acts on a 2.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t – 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 6.0 s.

a block of mass m=5 kg is pulled at a constant speed along a rough horizontal surface by a rope at 30 degree. the tension in the rope is T and the coefficient of kinetic friction between the block and the surface is 0.25. (a) find the tension in the rope. (b) if the block travels a distance of 4.5 m along the surface, what is the work done by the tension force from the rope? (c) find the work done by friction on the rock.

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.____ m

A particle is subject to a force F, that varies with position as shown in the following figure. F, (N) 3 1 х (m) 4 6 10 12 14 16 (a) Find the work done by the force on the particle as it moves from x = 0 to x = :3.00 m. (b) Find the work done by the force on the particle as it moves from x = 5.00 m to x = 9.00 m. J (c) Find the work done by the force on the particle as it moves from x = 11.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to X = 15.0 m?