Question 49: Laws of Motion

Question

A block of mass 1 kg , moving along $x$ with speed $v_i=10 \mathrm{~m} / \mathrm{s}$ enters a rough region ranging from $x=0.1 \mathrm{~m}$ to $x=1.9 \mathrm{~m}$. The retarding force acting on the block in this range is $\mathrm{F}_{\mathrm{r}}=-\mathrm{kr} \mathrm{N}$, with k $=10 \mathrm{~N} / \mathrm{m}$. Then the final speed of the block as it crosses rough region is.

Accepted Answer

Work Done = Final Kinetic Energy - Initial Kinetic Energy
$$W = \Delta K = K_f - K_i$$
Calculate the Work Done by the Retarding Force
The problem gives a retarding force $$F_r = -kr$$. It seems 'r' is a typo for 'x', the position, since the constant $$k$$ has units of N/m. So, the force is a function of position: $$F_r = -kx$$.
Because the force isn't constant, we need to find the work done by integrating the force over the distance the block travels in the rough region (from $$x = 0.1 	ext{ m}$$ to $$x = 1.9 	ext{ m}$$).
$$W = \int_{0.1}^{1.9} F_r \, dx = \int_{0.1}^{1.9} (-kx) \, dx$$
Plugging in $$k = 10 	ext{ N/m}$$:
$$W = -10 \int_{0.1}^{1.9} x \, dx$$
$$W = -10 \left[ \frac{x^2}{2} ight]_{0.1}^{1.9}$$
$$W = -5 \left[ (1.9)^2 - (0.1)^2 ight]$$
$$W = -5 [3.61 - 0.01] = -5(3.6)$$
$$W = -18 	ext{ J}$$
The negative sign just means the force did negative work, slowing the block down.
Calculate the Initial Kinetic Energy ($$K_i$$)
The initial kinetic energy is found using the formula $$K_i = \frac{1}{2}mv_i^2$$.
Given:
Mass $$m = 1 	ext{ kg}$$
Initial speed $$v_i = 10 	ext{ m/s}$$
$$K_i = \frac{1}{2}(1 	ext{ kg})(10 	ext{ m/s})^2 = \frac{1}{2}(100) = 50 	ext{ J}$$
Use the Work-Energy Theorem to find the Final Speed ($$v_f$$)
Now we can put everything together into the Work-Energy Theorem equation: $$W = K_f - K_i$$.
We know $$W = -18 	ext{ J}$$ and $$K_i = 50 	ext{ J}$$. The final kinetic energy is $$K_f = \frac{1}{2}mv_f^2$$.
$$-18 = \left(\frac{1}{2}(1)v_f^2ight) - 50$$
Now, let's solve for $$v_f$$:
$$-18 + 50 = \frac{1}{2}v_f^2$$
$$32 = \frac{1}{2}v_f^2$$
$$v_f^2 = 64$$
$$v_f = \sqrt{64}$$
$$v_f = 8 	ext{ m/s}$$
The final speed of the block as it leaves the rough region is
8 m/s
.
Therefore, the correct option is
C
.

Answer

$4 \mathrm{~m} / \mathrm{s}$

Answer

$10 \mathrm{~m} / \mathrm{s}$

Answer

$8 \mathrm{~m} / \mathrm{s}$

Answer

$6 \mathrm{~m} / \mathrm{s}$

Laws of Motion

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