A 3.5 kg block, initially in motion, is pushed along a horizontal floor by a force F of magnitude 19 N at an angle θ = 45° with the horizontal. The coefficient of kinetic friction between the block and floor is 0.25. (Assume the positive direction is to the right.)
And what, pray tell, is the question?
115.27N
To find the acceleration of the block, we can start by analyzing forces acting on it.
1. The force applied, F, can be resolved into its horizontal and vertical components. The horizontal component, F_x, is given by F * cos(θ) = 19 N * cos(45°) = 13.435 N.
2. The weight of the block, mg, acts vertically downward. The weight is given by m * g = 3.5 kg * 9.8 m/s^2 = 34.3 N.
3. The normal force, N, acts perpendicular to the surface and counterbalances the weight. Therefore, N = mg = 34.3 N.
4. The frictional force, F_friction, acts opposite to the direction of motion and is given by the equation F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. Therefore, F_friction = 0.25 * 34.3 N = 8.575 N.
Since the force of friction opposes the applied force, we need to subtract it from the horizontal component of the applied force to find the net force:
Net Force (F_net) = F_x - F_friction = 13.435 N - 8.575 N = 4.86 N.
Using Newton's second law, F_net = ma, we can solve for the acceleration (a):
4.86 N = (3.5 kg) * a.
Dividing both sides by 3.5 kg, we get:
a = 4.86 N / 3.5 kg = 1.39 m/s^2.
Therefore, the block has an acceleration of 1.39 m/s^2.