For the ball to accelerate upward, it is necessary that T? v m,g. In Equation (1), we replaced a, with a because the acceleration has only a y-component.For the block, we have chosen the x'-axis along the incline as in figure (c). For consistency with our choice for the ball, we choose the positive x'-direction to be down the incline.Apply the particle under a net force model to the block in the x'-direction and the particle in equilibrium model in the y'-direction. (For equation (2), use the following as necessary: g, m,, T, and 0. For equation (3), use the following as necessary: g, m,, n, and 0.)(2) EFx == m,ay = m,a(3) EFy== 0In Equation (2), we replaced a, with a because the two objects have accelerations of equal magnitude a.Solve Equation (1) for T. (Use the following as necessary: a, g, m,, and 0.)(4)T =Substitute this expression for T into Equation (2):m,g sin(8) – m,(g + a) = m,aSolve for a. (Use the following as necessary: m,, m2, g, and 8):(5)a =Substitute this expression for a into Equation (4) to find T. (Use the following as necessary: m,, m,, g, and 0):(6)T =Finalize The block accelerates down the incline only if m, sin(e) ? v m,. If m, ? v m, sin(8), the acceleration is up the incline for the block and downward for the ball. Also notice that the result for the acceleration, Equation (5), can be interpreted as themagnitude of the net external force acting on the ball-block system divided by the total mass of the system; this result is consistent with Newton's second law.EXERCISEConsider the following specific case of the system described in the example above. Suppose the angle e is 57.0° and the mass of the ball is 5.60 kg.Hint(a) Determine the mass of the block (in kg) if the tension in the cord is equal to the weight of the ball: T = m,g.kg(b) What is the acceleration in this case (in m/s2)? A ball of mass m, and a block of mass m, are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as in figure (a). The block lies on a frictionless incline of angle 8. Find the magnitude of the acceleration of the two objects and thetension in the cord.(c) The free-body diagram for the block. (The(a) Two objects connected by aincline is frictionless.)lightweight cord strung over a(b) The free-body diagramfrictionless pulley.for the ball.m2g sin emgm2g coseaSOLUTIONConceptualize Imagine the objects in the figure in motion. If m, moves down the incline, m, moves upward. Because the objects are connected by a cord (which we assume does not stretch), the magnitude of the acceleration of m, must be equal tothemagnitude of the acceleration of m,. Notice the normal coordinate axes in figure (b) for the ball and the "tilted" axes for the block in figure (c).Categorize We can identify forces on each of the two objects and we are looking for an acceleration, so we categorize the objects as particles under a net forcethe particle in equilibrium. For the block, this model is only valid for the x'-direction. In the y'-direction, we applyvV model because the block does not accelerate in that direction.Analyze Consider the free-body diagrams shown in figures (b) and (c).For which mass is there a normal force acting?O m,O m2O m, and m2O neither m, nor m2Apply Newton's second law in the y-direction to the ball, choosing the upward direction as positive. (Use the following as necessary: g, m,, T, and 0.)(1) EF,-|= m,ay = m,a

Question

For the ball to accelerate upward, it is necessary that T? v m,g. In Equation (1), we replaced a, with a because the acceleration has only a y-component.For the block, we have chosen the x'-axis along the incline as in figure (c). For consistency with our choice for the ball, we choose the positive x'-direction to be down the incline.Apply the particle under a net force model to the block in the x'-direction and the particle in equilibrium model in the y'-direction. (For equation (2), use the following as necessary: g, m,, T, and 0. For equation (3), use the following as necessary: g, m,, n, and 0.)(2) EFx == m,ay = m,a(3) EFy== 0In Equation (2), we replaced a, with a because the two objects have accelerations of equal magnitude a.Solve Equation (1) for T. (Use the following as necessary: a, g, m,, and 0.)(4)T =Substitute this expression for T into Equation (2):m,g sin(8) – m,(g + a) = m,aSolve for a. (Use the following as necessary: m,, m2, g, and 8):(5)a =Substitute this expression for a into Equation (4) to find T. (Use the following as necessary: m,, m,, g, and 0):(6)T =Finalize The block accelerates down the incline only if m, sin(e) ? v m,. If m, ? v m, sin(8), the acceleration is up the incline for the block and downward for the ball. Also notice that the result for the acceleration, Equation (5), can be interpreted as themagnitude of the net external force acting on the ball-block system divided by the total mass of the system; this result is consistent with Newton's second law.EXERCISEConsider the following specific case of the system described in the example above. Suppose the angle e is 57.0° and the mass of the ball is 5.60 kg.Hint(a) Determine the mass of the block (in kg) if the tension in the cord is equal to the weight of the ball: T = m,g.kg(b) What is the acceleration in this case (in m/s2)? A ball of mass m, and a block of mass m, are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as in figure (a). The block lies on a frictionless incline of angle 8. Find the magnitude of the acceleration of the two objects and thetension in the cord.(c) The free-body diagram for the block. (The(a) Two objects connected by aincline is frictionless.)lightweight cord strung over a(b) The free-body diagramfrictionless pulley.for the ball.m2g sin emgm2g coseaSOLUTIONConceptualize Imagine the objects in the figure in motion. If m, moves down the incline, m, moves upward. Because the objects are connected by a cord (which we assume does not stretch), the magnitude of the acceleration of m, must be equal tothemagnitude of the acceleration of m,. Notice the normal coordinate axes in figure (b) for the ball and the &#34;tilted&#34; axes for the block in figure (c).Categorize We can identify forces on each of the two objects and we are looking for an acceleration, so we categorize the objects as particles under a net forcethe particle in equilibrium. For the block, this model is only valid for the x'-direction. In the y'-direction, we applyvV model because the block does not accelerate in that direction.Analyze Consider the free-body diagrams shown in figures (b) and (c).For which mass is there a normal force acting?O m,O m2O m, and m2O neither m, nor m2Apply Newton's second law in the y-direction to the ball, choosing the upward direction as positive. (Use the following as necessary: g, m,, T, and 0.)(1) EF,-|= m,ay = m,a

Accepted Answer

From the Newtons second law on the ball,

For the ball to accelerate upwards, the necessary…