A 3.0 kg block, initially in motion, is pushed along a horizontal floor by a force F of magnitude 18 N at an angle = 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.) Calculate the magnitude of the frictional force on the block from the floor. Calculate the magnitude of the block's acceleration.
Break the force F in to vertical and horizontal components.
Friction= (mg + force vertical)mu
net force= ma
horizontal F - force friction = ma
Ok, so this is what I tried...
T = uk*m*g / ((cos theta)+uk(sin theta))
T = ((.25)(3)(9.8)) / ((cos -45)+(.25)(sin -45))
T = 13.9 N
then,
FNet = m*g - T*sin(theta)
FNet = (3)(9.8) - (13.9)*(sin -45)
FNet = 39.2
then,
fk = uk*FNet
fk = .25*39.2
fk = 9.8
Well that was the wrong answer but I don't see anything wrong with my approach. Any suggestions???
Your approach is mostly correct, but there is a mistake in calculating the frictional force. Let's go over the correct steps again.
1. Break the force F into horizontal and vertical components:
- The horizontal component F_horizontal = F * cos(theta) = 18 N * cos(45°) = 12.7 N
- The vertical component F_vertical = F * sin(theta) = 18 N * sin(45°) = 12.7 N
2. Calculate the net force acting on the block in the horizontal direction:
- F_net(horizontal) = F_horizontal - friction = 12.7 N - friction
3. Calculate the frictional force:
- The normal force N is equal to the weight of the block, N = m * g = 3 kg * 9.8 m/s^2 = 29.4 N
- friction = coefficient of kinetic friction * N = 0.25 * 29.4 N = 7.35 N
4. Substitute the frictional force value into the net force equation:
- F_net(horizontal) = 12.7 N - 7.35 N = 5.35 N
5. Calculate the acceleration of the block in the horizontal direction:
- F_net(horizontal) = m * a
- 5.35 N = 3 kg * a
- a = 5.35 N / 3 kg = 1.78 m/s^2
Therefore, the magnitude of the frictional force on the block from the floor is 7.35 N, and the magnitude of the block's acceleration is 1.78 m/s^2.
Note: Ensure that you use the correct signs for the components of forces and angles in your calculations. Also, double-check your trigonometric calculations using the appropriate units for angles (degrees or radians).