By force balance in the direction of inclined plane,
W*sin + Fk = m*a eq(1)
here, W = weight = m*g
m = mass of car = 40 kg
g = 9.81 /s^2
= angle of inclined plane = 37 deg
a = acceleration of car = ??
Fr = friction force applied on car = -*N
N = normal reaction force = ??
given, = Co-efficient of kinetic friction = 0.20
from force balance in the perpendicular direction of inclined plane,
N - W*cos = 0 (since there is no acceleration in the perpendicular direction of inclined plane.)
N = m*g*cos
So, from eq(1),
m*g*sin - *m*g*cos = m*a
a = g*sin - *g*cos
a= 9.81*sin(37 deg) - 0.20*9.81*cos(37 deg)
a = 4.34 m/s^2
"Let me know if you have any query."
A 40 kg box lies on a plane as shown below. If the coefficient of friction...
Problem 2:A 12-kg box is released from the top of an incline that is 5.0m long and makes an angle of 40° to the horizontal. A 60-N friction force impedes the motion of the box.a) What will be the acceleration of the box?b) How long will it take to reach the bottom of the incline?c) What is the coefficient of friction between box and incline?
In the following picture, a 12 kg box lies 10 m (measured along the plane of the ramp) from the bottom of an incline. What is the normal force acting on the box? How long will it take the box to slide down the ramp from its current position (assuming no friction)?
A block (M1) of 2.0 kg lying on an incline plane is connected to an equal mass (M2) by a massless cord passing over a pully, as shown below. If the coefficient of friction between M1 and the plane is 0.15; (a) determine the acceleration of M1 and M2 and (b) What minimum value of the coefficient of friction will keep the system from acceleration? 4. 30°
A 2 kg box sits on a ramp of 5 degrees where the coefficient of friction is 0.2. A 31 N force pulls the box uphill. Find the acceleration.
A 1.0 kg box starts from rest and slides down a very rough inclined plane as shown below. Use energy principles to determine the coefficient of friction on the incline surface if its speed at the bottom is 2.78 m/s. The box sits on a incline that is 0.5m above the ground on a 30 degree incline.
A 5 kg block is on a 52◦ inclined plane, where the coefficient of kinetic friction is µk = 0.46. A force of 30 N is applied along an axis parallel to the incline, as shown in the figure. The block is initially moving up the incline at 3 m/s. a) Draw a free-body diagram for the block. b) What is the acceleration of the block? c) How far does the block travel before it reaches its highest point? d)...
Two blocks are connected as shown below. The coefficient of kinetic friction between the block (M) and the incline is .19. Disregard any pulley mass of friction in the pulley. (A) calculate the magnitude of acceleration of the blocks. (B) find the tension in the string connecting the blocks. Use M=2.9kg. Please include the free body diagrams. 2M M 0:40
When in the position shown, the 5 kg box is moving down the inclined plane at a speed of 6m/s. What is the maximum force in the spring after the box hits it? The coefficient of kinetic friction between the box and the plane, is meu_k = 0.25, and the spring constant is k = 4 kN/m.
A 5kg box slides down an incline of angle 40 with an unknown coefficient of kinetic friction. If the box’s acceleration is measured to be 3.58 m/s2 down the slope, what is the value of the unknown coefficient of kinetic friction?
1. The force = 9 i + 12 j + 5 k acts on a small object. If the displacement of the object is = 16 i + 10 j . Find the work done by the force using the scalar product . 2. A 25 kg chandelier is hanging from a cable in an elevator (have you ever seen a chandelier in an elevator?). What is the tension in the cable if the elevator that moves downward...