To calculate the work required to push the block across the table at a steady speed, we start by using the formula for work when friction is involved.

Given dataTo calculate the work required to push the block across the table at a steady speed, we start by using the formula for work when friction is involved.

Given

• Mass of the block, $m = 12.0 \, \text{kg}$
• Coefficient of kinetic friction, $\mu_k = 0.60$
• Speed, $v = 1.00 \, \text{m/s}$
• Time, $t = 3.70 \, \text{s}$

Step 1: Calculate the Force of Kinetic Friction

The force of kinetic friction ($f_k$) acting on the block is given by: $f_k = \mu_k \cdot F_N$ where $F_N$ is the normal force. For a block on a horizontal table, the normal force equals the gravitational force: $F_N = m \cdot g$ where $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).

Substituting the values: $F_N = 12.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 117.6 \, \text{N}$ Then, $f_k = 0.60 \times 117.6 \, \text{N} = 70.56 \, \text{N}$

Step 2: Calculate the Distance Traveled

Since the block moves at a steady speed of $1.00 \, \text{m/s}$ for $3.70 \, \text{s}$, the distance $d$ it travels is: $d = v \cdot t = 1.00 \, \text{m/s} \times 3.70 \, \text{s} = 3.70 \, \text{m}$

Step 3: Calculate the Work Done

The work done $W$ to overcome the friction force is given by: $W = f_k \cdot d$ Substituting the values: $W = 70.56 \, \text{N} \times 3.70 \, \text{m} = 260.072 \, \text{J}$

Final Answer

The work required is approximately: $W \approx 260 \, \text{J}$

Would you like further explanation, or have any questions?

Here are five related questions you might find useful:

1. How would the work change if the coefficient of kinetic friction was higher?
2. What would happen if the mass of the block increased?
3. How would the work be affected if the block moved at a different constant speed?
4. How is work related to energy in this scenario?
5. What would the required work be if the surface were frictionless?

Tip: Remember that when moving an object at constant speed on a level surface, the applied force equals the frictional force.