A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.51. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x = 0. (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero. Figure P5.51
A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.51. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x = 0. (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero. Figure P5.51
A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.51. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x = 0. (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero.
Figure P5.51
(a)
Expert Solution
To determine
The acceleration of the block when x=0.400 m.
Answer to Problem 5.99CP
The acceleration of the block when x=0.400 m is 0.93m/s2.
Explanation of Solution
The mass of block is 2.20 kg, the tension T is 10 N, the height of pulley above the top of block is 0.100 m, and the coefficient of kinetic friction is 0.400.
Consider the free body diagram give below.
Figure I
Write the expression for the angle the block makes with the pulley
θ=tan−1(Hx) (I)
Here, θ is the angle that the pulley makes with the block, H is the vertical distance between the pulley and the block and x is the horizontal distance between the block and the pulley.
Substitute 0.400 m for x and 0.100 m for H to find θ.
θ=tan−1(0.100 m0.400 m)=14.03°
Write the formula to calculate normal reaction
N+Tsinθ=mgN=mg−Tsinθ
Here, N is the normal reaction, T is the tension, m is the mass of block and g is the acceleration due to gravity.
Substitute 2.20 kg for m, 10 N for T, 14.03° for θ and 9.8m/s2 for g in above equation to find N.
N=(2.20 kg)(9.8m/s2)−(10 N)sin(14.03°)=19.13 N
The equation for the force in x direction
Tcosθ−μN=ma (II)
Here, μ is the coefficient of kinetic friction and a is the acceleration.
Conclusion:
Substitute 2.20 kg for m, 10 N for T, 14.03° for θ, 0.400 for μ and 19.13 N for N in above equation to find a.
When x is very large, θ approaches zero and as x decreases, θ increases. The value of acceleration varies as (cosθ+(0.400)sinθ)−3.92. Thus, as x decreases, the acceleration increases initially, then drops quickly and becomes negative.
Conclusion:
Therefore, the acceleration increases initially then decreases and becomes negative as the value of x decreases.
(c)
Expert Solution
To determine
The maximum value of the acceleration and its corresponding position.
Answer to Problem 5.99CP
The maximum value of the acceleration is 0.975m/s2 which occurs at the position 0.250 m.
Explanation of Solution
Differentiate the equation (IV) with respect to θ in order to find the maximum value.
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