A 1.11 kg block slides down a 30.0 m long 28.0° incline at constant velocity. How much work is done by friction?
work=F.distance=force*distance*cosAngle between them
= mg*30*cos(90-28)
M*g = 1.11 * 9.8 = 10.9 N. = Wt. of block.
Fp = 10.9*sin 28 = 5.1 N. = Force parallel to incline.
Fn = 10.9*cos28 = 9.6 N. = Normal force.
u*Fn = Force of kinetic friction
Fp - u*Fn = M*a. a = 0(constant velocity).
5.1 - u*Fn = 1.11*0,
u*Fn = 5.1 N. = force of kinetic friction.
Work = 5.1 * 30 = 153 Joules.
To determine the work done by friction, we can use the formula:
Work = Force x Distance x cos(theta)
In this case, the force of friction opposes the motion of the block. Since it is moving at a constant velocity, we know that the force of friction is equal in magnitude and opposite in direction to the component of the gravitational force parallel to the incline.
First, let's calculate the gravitational force parallel to the incline. The force can be determined by multiplying the mass of the block by the acceleration due to gravity and the sine of the angle of the incline:
Force_gravity = mass x gravity x sin(theta)
Given:
mass = 1.11 kg
gravity = 9.8 m/s^2
theta = 28.0° (convert to radians)
θ (in radians) = (28.0° * π) / 180°
Next, calculate the force of gravity parallel to the incline:
Force_gravity = 1.11 kg x 9.8 m/s^2 x sin(28.0°)
Now, determine the friction force which is equal in magnitude to the gravitational force parallel to the incline:
Force_friction = Force_gravity
Finally, substitute the values into the work formula:
Work = Force_friction x Distance x cos(theta)
Given:
Distance = 30.0 m (length of the incline)
theta = 28.0° (convert to radians)
θ (in radians) = (28.0° * π) / 180°
Substituting the values:
Work = Force_friction x 30.0 m x cos(28.0°)
Calculating the value will give you the work done by friction.