The graph of the net force acting on the object of m = 3 (kg) along the x-axis is as shown in the figure. If the initial velocity of the object is 4 (m/s). Find its kinetic energy at x = 20 (m) in Joules. F, (N) x (m) 20 A) 204 В) 208 C) 218 D) 224 E) 236 20
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- A force F = (1.90 N )i + (1.00 N )j + (4.30 N )k acts on a 5.70 kg mobile object that moves from an initial position of (8.90 m )i + (3.60 m )j + (5.40 m )k in 4.70 s. Find (a) the work done on the object by the force in the 4.70 s interval, (b) the average power due to the force during that i = (5.30 m )i + (6.30 m )j + (2.70 m )k to a final position of d f = interval, and (c) the angle between vectors d ¡ and d f. (a) Number i Units (b) Number Units (c) Number i UnitsA 3.35-kg particle is subject to a net force that varies with position as shown in the figure. The particle starts from rest at x = 0. F. (N) 3 2 1 0 I 2 4 6 8 10 12 14 16 What is its speed at the following positions? (a) x = 5.00 m (b) x = 10.0 m (c) x = 15.0 m Submit Answer m/s m/s m/s Need Help? Read It Watch It x (m)4. Here we prove a version of the work-kinetic energy theorem for an object moving in 1-D subject to a constant force. Consider a mass m, traveling initially at speed v, on which a constant net force F is applied in the same direction as its motion over a distance d. After the net force has acted over this distance, the speed of the object is v2. In terms of m, F, v1, v2, and d only, V, A) What is the acceleration of the mass? t=0 m F d t>0_m B) What is the time over which the mass travels the V, distance d? [Use one equation of constant acceleration to find this.] C) Use another equation of constant acceleration to relate m, F, v,, v2, and d. Show that the product W = Fd is equal to the change in the mass’s kinetic energy.
- An object of mass 5.00 kg is subject to a force Fx that varies with the position as in the figure below. (a) Find the work done by the force on the object as it moves from x = 0 to x = 16.00 m. (b) Find the speed of the object at x=16 m, if the object was initially at rest. F(N) 4 r(m) 10 12 14 16 18 20Ex(N) 80 50 25 X(m) 10 25 40 50 58 1) Work by a force, drawn on a graph, determine: c) Kinetic energy at position x = 58 m, if at the beginning of position x = 0 the velocity of the object is zero? d) Velocity at x=25m if the mass of the object is 5 kgThe displacement vector from the grocery store door to your car is ∆x = (5 i – 8 j ) m but because of the strong gust, you have to apply a force F in a slightly different direction to keep your cart moving straight from the door to the car, F = (30 i + 4π j ) N. How much work do you do? a) 306.8 J b) 49.5 J c) 250.5 J
- A 3.3 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F(x) = (7.4-x²) N. where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x 3.8 m? (b) What is the maximum kinetic energy of the block between x=0 and x 3.8 m? (a) Number (b) Number Units UnitsAn object whose mass is 10 lb is projected straight upward from the surface of the earth and reaches a height of 100 ft when its velocity reaches zero. The only force acting on the object is the force of gravity. The acceleration of gravity is g = 32.2 ft/s². Determine the initial kinetic energy of the object, in ft·lbf, and the initial velocity of the object, in ft/s.