A horizontal force F is exerted on a 20-kg box to slide it up a 30 degrees incline. The friction force retarding the motion is 80 N.How large must F be if the acceleration of the moving box is to (a)zero and (b) 0.75m/s2?
(a) 206 N
(b) 233 N
To find the required force, we need to consider the forces acting on the box.
Let's begin with the forces acting on the box on the incline:
1. The weight (mg): The weight of the box is given by the formula mg, where m is the mass of the box (20 kg) and g is the acceleration due to gravity (9.8 m/s^2). So, the weight is 20 kg × 9.8 m/s^2 = 196 N.
2. The component of weight acting parallel to the incline (mg sinθ): The weight vector can be split into two components – one perpendicular to the incline and one parallel to the incline. The component parallel to the incline is given by mg sinθ, where θ is the angle of inclination (30 degrees). So, mg sinθ = 20 kg × 9.8 m/s^2 × sin(30 degrees) = 98 N.
3. The friction force (f): The given friction force retarding the motion is stated as 80 N.
Now, let's consider the forces in different scenarios:
(a) To find the force required to bring the box to zero acceleration:
In this case, the box is not accelerating, meaning the net force is zero. The net force on the box is given by the equation:
Net Force = Applied Force (F) - Friction Force (f) - Component of weight parallel to the incline (mg sinθ)
Since the box is not accelerating, the net force is zero. So, we have:
0 = F - f - mg sinθ
Substituting the known values, we get:
0 = F - 80 N - 98 N
Rearranging the equation, we find:
F = 80 N + 98 N
F = 178 N
Therefore, the force required to bring the box to zero acceleration is 178 N.
(b) To find the force required for an acceleration of 0.75 m/s^2:
In this case, the net force on the box is given by the equation:
Net Force = Applied Force (F) - Friction Force (f) - Component of weight parallel to the incline (mg sinθ)
Since the box is accelerating with an acceleration of 0.75 m/s^2, the net force is given by:
Net Force = mass (m) × acceleration (a)
So, we have:
m × a = F - f - mg sinθ
Substituting the known values, we get:
20 kg × 0.75 m/s^2 = F - 80 N - 98 N
Rearranging the equation, we find:
F = 20 kg × 0.75 m/s^2 + 80 N + 98 N
F = 15 N + 178 N
F = 193 N
Therefore, the force required for an acceleration of 0.75 m/s^2 is 193 N.
Next step of queßtion
Phy
weight force down slope = m g sin 30
= .5 m g
F force up slope = F cos 30 = .866 F
friction fore down slope = 80
total force up slope = .866 F -.5mg -80
= m a