# A 3.0 kg block released from the top of a rough ramp accelerates down the ramp at 1.6m/s^2.If the force of kinetic friction on the block is 10 N, what is the angle of the ramp?

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### 1 Answer

When the block is placed on the ramp it experiences a gravitational force of attraction acting vertically downwards. This can be divided into two components, one along the ramp and the other normal to the ramp.

The gravitational force is given by m*g, where m is the mass of the block and g is the acceleration due to gravity.

If the angle made by the ramp with respect to the horizontal is x, the component along the ramp is m*g*sin x

The resistive force due to friction is 10 N. This gives the net force that leads to the acceleration of the block downwards as m*g*sin x - 10

As the block accelerates down the ramp at 1.6 m/s^2

m*g*sin x - 10 = m*(1.6)

=> 3*9.8*sin x - 10 = 3*1.6

=> sin x = (3*1.6 + 10)/3*9.8

=> sin x = 0.5034

x = arc sin 0.5034

=> x = 30.22 degrees.

**The angle of the ramp is 30.22 degrees.**