Question

A 2.0 kg steel block is at rest on a steel table. A horizontal string pulls on the block.
(a) What is the minimum string tension needed to move the block?
(b) If the string tension is 20 N, what is the block’s speed after moving 1.0 m?
(c) If the string tension is 20 N and the table is coated with oil, what is the block’s speed
after moving 1.0 m?

Answer 1

Answer on Question#45361, Physics, Mechanics | Kinematics | Dynamics

a) Since the body is at rest at the table, the normal force is equal to gravitational force ($N = mg$). In order for block to move, the tension of the string must be bigger than the frictional force, which is $F_f = \mu N = \mu mg$, where $\mu$ is the coefficient of static friction.

For clean steel, this coefficient is equal to $\mu_1 = 0.74$. Hence, minimum string tension is $T = F_f = \mu mg \approx 14.52\,N$.

b) First, let us find the acceleration of the block, if string tension is $20\,N$. According to $2^{\text{nd}}$ Newtons law, $a = \frac{F_{net}}{m} = \frac{F - \mu mg}{m} = \frac{F}{m} - \mu g \approx 2.74\, \frac{m}{s^2}$.

Hence, velocity of the block as a function of time is $v(t) = at$. The displacement is $S(t) = \frac{at^2}{2}$. The time needed to move $S$ meters is $t = \sqrt{2\, \frac{S}{a}}$, thus velocity at that moment is $v = a \sqrt{2\, \frac{S}{a}} = \sqrt{2\, S a} \approx 2.34\, \frac{m}{s}$.

c) Using formulas from $b)$ with coefficient of friction of lubricated steel $\mu_2 = 0.16$, obtain $a = 8.43\, \frac{m}{s^2}$ and $v = 4.1\, \frac{m}{s^2}$.

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