Calculation of Resulting Acceleration for a Block
Calculation of Resulting Acceleration for a Block
The magnitude of each force is 208 N, the force on the right is applied at an angle 36°, and the mass of the block is 17 kg. The coefficient of friction is 0.149 and the acceleration of gravity is 9.8 m/s^2.
What is the magnitude of the resulting acceleration?
What is the magnitude of the resulting acceleration?
Answer:
11.06m/s²
Explanation:
According to Newton's second law of motion:
\[F_x = ma_x\] \[F_m - F_f = ma_x\] \[mg\sin\theta - \mu R mg\cos\theta = ma_x\]Given:
Mass \(m = 17kg\)
Force \(F_m = 208N\)
Angle \(\theta = 36\) degrees
Acceleration due to gravity \(g = 9.8 m/s^2\)
Let \(a\) be the acceleration.
Substitute:
\(208 - 0.149(17)(9.8)\cos 36 = 17a\)
\(208 - 24.6568\cos36 = 17a\)
\(208 - 19.9478 = 17a\)
\(188.05 = 17a\)
\(a = 188.05/17\)
\(a = 11.06 m/s^2\)
Therefore, the magnitude of the resulting acceleration is 11.06 m/s^2.