Question

The figure below shows a block with mass m = 5.3 kg pulled up a ramp inclined at an angle of 6 = 28° with a force of magnitude F = 33 N parallel to the ramp. m (a) If there is no friction between the block and the ramp, what is the magnitude of the block's acceleration (in m/s2)? 1.485 Draw a free-body diagram. What forces act on the block? Let the x-axis be parallel to the incline. What is the direction of the force F? What is its x-component? What is the direction of the gravitational force? What is its x-component? Can you use Newton's second law and the x-components of the forces to find the acceleration? m/s2 (b) If instead the coefficient of kinetic friction between the block and the ramp is 0.12, what is the magnitude of the block's acceleration in m/s2)? 49 X What is the direction of the frictional force? What is the direction of the normal force? How does the frictional force depend on the normal force? Let the x-axis be parallel to the incline and the y-axis rise from the incline at a right angle. By Newton's second law, if there is no acceleration in the y-direction (perpendicular to the ramp), what must be the magnitude of the normal force? Use Newton's second law and the x-components of the forces to find the acceleration. m/s2

Answer 1

solution, m=5. 5.3 kg, F = 33 N, O = 28° (a) N F Net Force cep the inclined plane Fnet = F- F-masine ma= F - mgsine a= F g sino mg a = - 9.8 x sin 28 = 1.62559 m/s mgsino mgcoso 33 5.3 a= (6) N uk Nuk mgsino 1.62559 m/s2 Now friction will act down the ramb N = mg cose, SO ME 0.12 fx= F-fk- mg sino F Mk mgcos e-mgsino a = F/m- uk ung aso-gsino FR mgcse = ma = ma 33 -0.12 X9.8x asko 5.3 -9.8xsın2.80 a=0.58725 m/s²