A 10-kg box is pulled along a horizontal surface by a
force of 40N applied at a 30° angle above
horizontal. The coefficient of kinetic friction is 0.3.
Calculate the acceleration.
Wb = M*g = 10 * 9.8 = 98 N.
Fn = 98-40*sin30 = Normal force.
Fk = u*Fn = Force of kinetic friction.
a = (Fap*Cos30-Fk)/M.
A box of mass 10kg is at rest on a horizontal plane.A force of 20N is applied to it at 30 degrees to the plane.assuming the is no frictional force between the box and the plane,calculate the acceleration of the box.
2m/s
To calculate the acceleration, we need to break down the forces acting on the box and apply Newton's second law of motion.
1. Determine the vertical and horizontal components of the applied force:
The vertical component of the force can be calculated using the formula:
Vertical component = Force * sin(angle)
Vertical component = 40 N * sin(30°) = 20 N
The horizontal component of the force can be calculated using the formula:
Horizontal component = Force * cos(angle)
Horizontal component = 40 N * cos(30°) = 34.64 N
2. Calculate the force of kinetic friction:
The force of kinetic friction can be calculated using the formula:
Force of friction = coefficient of friction * normal force
Since the box is on a horizontal surface, the normal force is equal to the weight of the box, which is the mass * acceleration due to gravity:
Normal force = mass * acceleration due to gravity
Normal force = 10 kg * 9.8 m/s^2 = 98 N
Therefore, the force of kinetic friction is:
Force of friction = 0.3 * 98 N = 29.4 N
3. Calculate the net force acting on the box in the horizontal direction:
Net force = Horizontal component of applied force - Force of friction
Net force = 34.64 N - 29.4 N = 5.24 N
4. Calculate the acceleration using Newton's second law of motion:
Net force = mass * acceleration
5.24 N = 10 kg * acceleration
Rearranging the equation, we find:
acceleration = Net force / mass
acceleration = 5.24 N / 10 kg = 0.524 m/s^2
Therefore, the acceleration of the box is 0.524 m/s^2.