Physics, 19.12.2020 14:00 TheOneHpLeft

A block with mass m = 0.200 kg is placed against a compressed spring at the bottom of a ramp that is at an angle of 53.0∘ above the horizontal. The spring has 8.00 J of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of 3.00 m, its speed is 4.00 m/s. Part A What is the magnitude of the friction force that the ramp exerts on the block while the block is moving? Express your answer with the appropriate units.

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Answer from: batmannn1516

$0.566\; \rm N$ (assuming that while the block is moving, the friction on the block is constant.)

Explanation:

The mechanical energy of a system is the sum of its:

elastic potential energy, gravitational potential energy, andkinetic energy.

Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.

Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: $8\; \rm J$ according to the question.

The question states that the ramp is $53.0^\circ$ above horizontal. Therefore, after the block has traveled $3.00\; \rm m$ (along the ramp,) the height of this block would have increased by $\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m$ . Calculate the corresponding gain in gravitational potential energy:

$\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}$ .

On the other hand, the question states that the speed of the block ( $m = 0.200\; \rm kg$ ) at that moment is $v = 4.00\; \rm m \cdot s^{-1}$ . Calculate the corresponding kinetic energy:

$\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}$ .

Because the block was already released, there should be no elastic potential energy in the spring.

Hence, the mechanical of the block-spring system would be approximately $4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J$ .

Approximate the amount of mechanical energy that is lost:

$8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J$ .

In other words, when applied over $3\; \rm m$ , the friction on this block would do approximately $1.69923\; \rm J$ of work. Approximate the size of that friction:

$\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}$ .

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