Question

Figure shows a spring attached to a 2.0 kg block. The other end of the spring is pulled by a motorized toy train that moves forward at constant speed 6.0 cm/s. The spring constant is 50 N/m and the coefficient of static friction between the block and the surface is 0.60. The spring is at its natural length at t=0s when the train starts to move. When does the block slip (express your answer in second)?

Answer 1

4.0

Answer 2

The problem describes a block attached to a spring, with the other end of the spring being pulled by a toy train moving at a constant speed. The block rests on a surface with friction. We need to find the time at which the block begins to slip.

1. Condition for the block to slip: The block will start to slip when the force exerted by the spring ($F_{spring}$) on the block exceeds the maximum static friction force ($f_{s,max}$) between the block and the surface. Until this point, the block remains stationary.

2. Calculate the maximum static friction force ($f_{s,max}$): The normal force ($N$) acting on the block is equal to its weight ($mg$) since the surface is horizontal. Given: Mass of the block, $m = 2.0$ kg Coefficient of static friction, $\mu_s = 0.60$ We will use the standard acceleration due to gravity, $g = 10 \text{ m/s}^2$ for simplicity in calculations, which is common in such problems unless specified otherwise.

The maximum static friction force is: $f_{s,max} = \mu_s N = \mu_s mg$ $f_{s,max} = 0.60 \times 2.0 \text{ kg} \times 10 \text{ m/s}^2$ $f_{s,max} = 12 \text{ N}$

3. Determine the spring extension ($x$) when the block slips: According to Hooke's Law, the force exerted by the spring is $F_{spring} = kx$, where $k$ is the spring constant and $x$ is the extension of the spring from its natural length. Given: Spring constant, $k = 50 \text{ N/m}$

When the block is about to slip, the spring force equals the maximum static friction force: $F_{spring} = f_{s,max}$ $kx = f_{s,max}$ $50 \text{ N/m} \times x = 12 \text{ N}$ $x = \frac{12 \text{ N}}{50 \text{ N/m}}$ $x = 0.24 \text{ m}$

4. Calculate the time ($t$) for this extension to occur: The train moves at a constant speed, $v = 6.0 \text{ cm/s}$. Convert this to meters per second: $v = 6.0 \text{ cm/s} = 0.06 \text{ m/s}$

Since the spring is at its natural length at $t=0$s and the block remains stationary until it slips, the extension of the spring is equal to the distance moved by the train. Distance moved by train = $vt$ So, the extension $x = vt$ $0.24 \text{ m} = (0.06 \text{ m/s}) \times t$ $t = \frac{0.24 \text{ m}}{0.06 \text{ m/s}}$ $t = 4.0 \text{ s}$

The block will slip after 4.0 seconds.