A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s² up the incline. Determine the magnitude of the force of friction acting on the block. A) 1.9 N B) 2.2 N C) 1.3 N D) 1.6 N E) 3.3 N

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Quizplus · Accepted Answer

To find the force of friction, we first calculate the net force acting on the block using Newton's second law, $F_{net} = ma$, where $m = 1.0 \, \text{kg}$ and $a = 1.4 \, \text{m/s}^2$. The net force is $F_{net} = 1.0 \, \text{kg} \times 1.4 \, \text{m/s}^2 = 1.4 \, \text{N}$. This net force is the result of the applied force minus the force of friction and any component of gravitational force along the incline. However, since we're given the acceleration and the applied force, we focus on these to find the frictional force.The applied force is 7.0 N, and for the block to accelerate at $1.4 \, \text{m/s}^2$, the difference between the applied force and the frictional force must equal the net force. Thus, $7.0 \, \text{N} - F_{friction} = 1.4 \, \text{N}$. Solving for $F_{friction}$, we get $F_{friction} = 7.0 \, \text{N} - 1.4 \, \text{N} = 5.6 \, \text{N}$.However, this calculation does not account for the gravitational force component along the incline, which also opposes the applied force. To correct this oversight, we calculate the gravitational force component along the incline: $F_{gravity} = mg\sin(\theta) = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(22^\circ) \approx 3.7 \, \text{N}$.The actual net force that causes the acceleration is the result of the applied force minus both the frictional force and the gravitational component: $7.0 \, \text{N} - F_{friction} - 3.7 \, \text{N} = 1.4 \, \text{N}$. Rearranging to solve for $F_{friction}$, we find $F_{friction} = 7.0 \, \text{N} - 1.4 \, \text{N} - 3.7 \, \text{N} = 1.9 \, \text{N}$. Therefore, the correct magnitude of the force of friction acting on the block is 1.9 N.

A 10-Kg Block Is Pushed Up a Rough 22° Inclined Plane