Question A 10kg crate is placed on a horizontal conveyor belt. The materials are such that μs=0.5 and μk=0.3 A. Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. b.What is the maximum acceleration the belt can have without the crate slipping

Step1/3(a)The free body diagram showing all the forces on crate if the conveyer belt is speeding up is shown below:Here, n is the normal force, fs​ is the static frictional force, FNet​ is the net force, and FG​ is the force due to gravity.Explanation:The free body diagram is drawn using the direction of the forces acting on the crate. The frictional force is always opposite to the direction of motion. The normal force is always opposite to the direction of gravity force.Step2/3(b)Calculate the maximum acceleration the belt can have without the crate slipping.The net force on the belt is expressed as follows: F=−μs​mg Here, μs​ is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.Substitute 10 kg for m, 9.8m/s2 for g, and 0.5 for μs​ in expression F=−μs​mg . F=−(0.5)(10kg)(9.8m/s2)=−49N​ Explanation:The net force on the crate is equal to the static frictional force on the crate. The negative sign for static frictional force on the crate represents the direction of force is opposite to the direction of motion of crate.Step3/3The Newton’s second law states that the net force acting on the object is directly proportional to the acceleration. F=ma Here, m is the mass, a is the acceleration, and F is the force.Substitute -49 N for F in expression F=ma . −49N=ma Substitute 10 kg for m in expression −49N=ma . −49N=(10kg)aa=10kg−49N​=−4.9m/s2​ The magnitude of the maximum acceleration of the belt is 4.9m/s2 Explanation:According to Newton’s second law of motion, the net force on an object is equal to product of mass m of the object and acceleration of the object a.The acceleration is calculated using the fact that the force acting on the belt is weight force and the frictional force. ...