Answer:
The force due to friction on the cart is 245 N.
Explanation:
Given that,
Mass of the cart, m = 50 kg
It is placed on an incline of 30 degrees above the horizontal.
Initially, the cart is at rest.
The coefficient of static friction is 0.85 and the coefficient of kinetic friction is 0.5.
The net force acting along the horizontal direction is given by :
[tex]F=f-mg\sin \theta[/tex]
Here, net force is 0.
[tex]f=mg\sin \theta\\\\f=50\times 9.8\times \sin (30)\\\\f=245\ N[/tex]
So, the force due to friction on the cart is 245 N.
The force due to friction on the 50 kg cart which is inclined at 30 degree will be "245 N".
According to the question,
Cart's mass, m = 50 kg
Inclined angle, θ = 30°
Coefficient of static friction = 0.85
Coefficient of Kinetic friction = 0.50
Acceleration due to gravity, g = 9.8 m/s²
We know,
The net force be:
→ F = f - mgSinθ
or,
f = mgSinθ
By substituting the values, we get
= 50 × 9.8 × Sin30°
= 245 N
Thus the above approach is right.
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