The force acting on a 3.00-kg particle is F,=8- x+16 N. (a) Find the net work done by this force on the particle as it moves from x=0 to x=3.00 m. (b) Assuming the particle started from rest at x=0, find its speed at x=5.00 m.
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- A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0◦ with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by gravity? (2pts) (b) How much work is done by the 100 N force?A skier of mass 107 kg travels down a frictionless ski trail. (a) If the top of the trail is a height 186 m above the bottom, what is the work done by gravity on the skier? J (b) Find the velocity of the skier when he reaches the bottom of the ski trail. Assume he starts from rest. m/sA particle is subject to a force Fx that varies with position as shown. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 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?
- A Net Force Fnet=(x-3x2)i + (2y2-y3)j acts on a 3.5 kg particle. a) Calculate the net work done on the particle as it moves from (1.0m,0.0m) to (1.0m,4.0m). b) At (x,y) = (1.0m,0.0m), the particle's speed is 5.0 m/s. Find the particle's speed at (x,y) = (1.0m,4.0m).A person pulls a 75-kg box 20 m along a horizontal floor by a constant force Fp = 125 N, which acts at a 42 degree angle. The floor is not smooth and exerts a friction force of Ff = 65 N. Determine the following: The work done by each force acting on the crate (Don’t forget any!) The net work done on the crateA single conservative force acts on a 4.70-kg particle within a system due to its interaction with the rest of the system. The equation Fx = 2x + 4 describes the force, where Fx is in newtons and x is in meters. As the particle moves along the x axis from x = 0.96 m to x = 7.50 m, calculate the following. (a) the work done by this force on the particle 81.488 J (b) the change in the potential energy of the system J (c) the kinetic energy the particle has at x = 7.50 m if its speed is 3.00 m/s at x = 0.96 m J
- A mass m is acted on by a force F,(x) = Ax' from x = 0 to x = L. (a) If m starts from rest, what is its speed at x = L? (b) Beyond x = L, F, = 0, but the mass encounters a surface with coefficient of kinetic friction 4. How far does m go before coming to rest? [Find the answer using the work-KE theorem.](a) A force F= (2x1 + 4yĵ), where F is in newtons and x and y are in meters, acts on an object as the object moves in the /F. F. dr done by the force on the object (in J). x-direction from the origin to x = 4.63 m. Find the work W = J (b) What If? Find the work W = (4.63 m, 4.63 m) along a straight-line path making an angle of 45.0° with the positive x-axis. J F. dr done by the force on the object (in J) if it moves from the origin to Is the work done by this force dependent on the path taken between the initial and final points? dependent independentThe force on a particle is directed along an x axis and given by F = Fo(x/xo - 1) where x is in meters and Fis in Newtons. If Fo= 2.1 N and xo = 2.0 m, find the work done by the force in moving the particle from x = 0 to x = 2x0 m.
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