Question

A block is allowed to slide down a frictionless plane which is inclined at 42 degree with the horizontal. A force of 56. 0 N is applied parallel and toward the top of the plane. If the block has a mass of 15.0 kg, what is the kinetic energy of the block when it has moved 2.50m down the plane?

Answer 1

Answer on Question #43396, Physics, Mechanics | Kinematics | Dynamics

A block is allowed to slide down a frictionless plane which is inclined at 42 degree with the horizontal. A force of $56.0\mathrm{N}$ is applied parallel and toward the top of the plane. If the block has a mass of $15.0\mathrm{kg}$ , what is the kinetic energy of the block when it has moved $2.50\mathrm{m}$ down the plane?

Solution.

From the law of conservation of energy:

$$E _ {1} + A _ {n c} = E _ {2}$$

Where $A_{nc}$ is a work of non-conservative forces. In our case this force is $\mathbf{F}$ . So:

$$E _ {p o t 1} + E _ {K 1} + (- F \cdot l) = E _ {p o t 2} + E _ {K 2}$$

Work of force is $\mathbf{F}$ have is negative because block is moving towards opposite direction related to vector $\mathbf{F}$ . Initially block was at the state of rest. So:

$$m g h _ {1} + 0 - F \cdot l = m g h _ {2} + E _ {K 2}$$ $$E _ {K 2} = m g \left(h _ {1} - h _ {2}\right) - F \cdot l = m g l \cdot \sin (\alpha) - F \cdot l = \left(m g \sin (\alpha) - F\right) \cdot l$$

Numerically:

$$E _ {K 2} = 1 5 k g \cdot 9. 8 \frac {m}{s ^ {2}} \cdot 2. 5 m \cdot \sin (4 2 {}^ {\circ}) - 5 6 N \cdot 2. 5 m \approx 1 0 6 J$$

Answer: 106 J

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