Question

a dockworker applies a constant horizontal force of 88.0 n to a block of ice on a smooth horizontal floor. the frictional force is negligible. the block starts from rest and moves a distance 13.0 m in a time of 4.60 s. you may want to review (page). for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of determining force from acceleration. part b if the worker stops pushing after 4.60 s, how far does the block move in the next 5.00 s? express your answer in meters. view available hint(s) x = \boxed{} m

Explanation:

Step1: Find acceleration using \( x = v_0t + \frac{1}{2}at^2 \)

Given \( v_0 = 0 \, \text{m/s} \), \( x = 13.0 \, \text{m} \), \( t = 4.60 \, \text{s} \).
\( 13.0 = 0 + \frac{1}{2}a(4.60)^2 \)
\( a = \frac{2 \times 13.0}{(4.60)^2} \approx 1.23 \, \text{m/s}^2 \)

Step2: Find velocity at \( t = 4.60 \, \text{s} \) using \( v = v_0 + at \)

\( v = 0 + 1.23 \times 4.60 \approx 5.66 \, \text{m/s} \)

Step3: Find distance in next \( 5.00 \, \text{s} \) (constant velocity, \( a = 0 \))

Using \( x = vt \), \( x = 5.66 \times 5.00 \approx 28.3 \, \text{m} \)

Answer:

\( \approx 28.3 \)