A 225-kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.
1.2 m/s^2
To calculate the acceleration of the crate, we need to consider the force applied and the force of friction acting on the crate.
First, let's calculate the force of friction using the coefficient of friction. The force of friction can be calculated using the equation:
Force of Friction = Coefficient of Friction × Normal Force
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the crate is pushed horizontally, so the normal force is equal to the weight of the crate, given by:
Normal Force = Mass × Gravity
The mass of the crate is given as 225 kg and the acceleration due to gravity is approximately 9.8 m/s^2.
Normal Force = 225 kg × 9.8 m/s^2
Next, we can find the force of friction:
Force of Friction = 0.20 × Normal Force
Now, let's find the net force acting on the crate. The net force is the difference between the applied force and the force of friction. Since the crate is being pushed horizontally, the frictional force acts in the opposite direction of the applied force:
Net Force = Applied Force - Force of Friction
Now, we can calculate the acceleration using Newton's second law of motion:
Net Force = Mass × Acceleration
Rearranging the equation:
Acceleration = Net Force / Mass
Plugging in the values we have:
Acceleration = (Applied Force - Force of Friction) / Mass
Now, let's substitute the values we found:
Acceleration = (710 N - (0.20 × 225 kg × 9.8 m/s^2)) / 225 kg
Calculating the force of friction:
Force of Friction = 0.20 × (225 kg × 9.8 m/s^2)
Substituting the values:
Acceleration = (710 N - (0.20 × 225 kg × 9.8 m/s^2)) / 225 kg
Simplifying the equation gives us the acceleration of the crate.
A 225-kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate
i dont kbow
net force= mass*acceleration
710-mu(mg)=ma
solve for a.