Question

A 2kg block is sliding up a surface inclined at an angle of 30° above the horizontal. The block has the speeds indicated in the figure when it passes points L and K. The distance between points L and K is 2m. Which of the following includes a correct claim about the interaction of the block and the surface with supporting evidence from the figure?

Answer 1

Answer:

The surface is rough, because the work-energy theorem indicates the block should be traveling faster at point  K .

Explanation:

Correct. Using the work-energy theorem, the speed at point K would be U1+K1=U2+K2

0+1/2mv^21=mgh2+1/2mv^22

v2= square root v^21−2gh2

=square root (7ms)^2−(2)(10)(2m)(sin30)

v2=5.4ms.

Since the actual speed at point K is slower than this value, there must have been friction between the block and surface.

Answer 2

The block is decelerating at the rate of 9.84 m/s² as it slides up the inclined plane.

The given parameters;

• mass of the block, m = 2 kg
• angle of inclination of the block, θ = 30⁰
• distance traveled by the block, = 2m

The normal force on the block is calculated as

Fₙ = mgcosθ

Fₙ = (2 x 9.8) cos(30)

Fₙ = 16.97

The coefficient of kinetic friction between the block and the surface;

μ = tanθ

μ = tan(30)

μ = 0.58

The frictional force on the block along the surface is calculated as;

$F_k = \mu F_n\\\\F_k = 0.58 \times 16.97\\\\F_k = 9.84 \ N$

The net horizontal force on the block as it slides up is calculated as;

$\Sigma F_x = 0\\\\-mgsin(\theta) - F_k = ma\\\\-(2 \times 9.8 \times sin(30) ) - 9.84 = 2a\\\\-19.64 = 2a\\\\a = -9.84 \ m/s^2$

Thus, we can conclude that the block is decelerating at the rate of 9.84 m/s² as it slides up the inclined plane.

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