A 10-KG block is pulled along a rough horizontal floor by a 12-N force acting at a 30 degrees angle with respect to the horizontal. The block accelerates at 0.1 m/s2. Calculate
A....... the weight of the block
B...... the force of friction acting on the block
C.... the normal force acting in the block
d.........the coefficient of kinetic friction uk
e..... the force of static friction acting on the block if the pulling force must be 20 N acting at 30 degrees in other start the block moving along the floor
F.... the coefficient of static friction Us
This question appeared yesterday under a different name. My answer remains the same.
weight of the block is 98
force of friction acting on the block is 1N
To solve for the various quantities, we'll break down the problem and use Newton's second law, F=ma, along with some trigonometry.
A. The weight of the block can be calculated using the formula weight = mass * acceleration due to gravity. The mass of the block is given as 10 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So, weight = 10 kg * 9.8 m/s^2 = 98 N.
B. To find the force of friction acting on the block, we'll first calculate the net force acting on the block. The net force is the vector sum of the applied force and the force of friction. Since the block is accelerating, we know that the net force is equal to mass times acceleration (F_net = m * a). Therefore, F_net = 10 kg * 0.1 m/s^2 = 1 N.
Next, we need to find the horizontal and vertical components of the applied force. The horizontal component can be calculated as F_h = F_applied * cos(theta), where theta is the angle with respect to the horizontal. So, F_h = 12 N * cos(30 degrees) = 10.39 N.
The force of friction acts in the opposite direction of the net force, so it is equal in magnitude but opposite in direction. Thus, the force of friction acting on the block is 1 N.
C. The normal force acting on the block is the force exerted by the surface perpendicular to the block. In this case, since the block is on a horizontal floor and is not moving in the vertical direction, the normal force is equal in magnitude but opposite in direction to the weight of the block. Therefore, the normal force is 98 N.
D. The coefficient of kinetic friction (uk) can be found using the formula F_friction = uk * N, where N is the normal force. We already know that the force of friction is 1 N, and the normal force is 98 N. So, uk = F_friction / N = 1 N / 98 N = 0.01.
E. When the pulling force must be 20 N to start the block moving along the floor, we'll calculate the force of static friction acting on the block. The force of static friction is equal in magnitude but opposite in direction to the applied force, up to a certain maximum value. Since the block is just about to start moving, the force of static friction is at its maximum value.
Using the formula for the maximum force of static friction (F_static = Us * N), we can rearrange it to solve for Us: Us = F_static / N. Plugging in the values, Us = 20 N / 98 N = 0.204.
F. The coefficient of static friction (Us) is calculated in question E, and is equal to 0.204.