Question 2.4: Lifting a suitcase Earth exerts a downward 100-N force on a ......
Lifting a suitcase
Earth exerts a downward 100-N force on a 10-kg suitcase. Suppose you exert an upward 120-N force on the suitcase. If the suitcase starts at rest, how fast is it traveling after lifting for 0.50 s?
Sketch and translate First, we make a sketch of the initial and final states of the process, choosing the suitcase as the system object. The sketch helps us visualize the process and also brings together all the known information, letting our brains focus on other aspects of solving the problem. One common aspect of problems like this is the use of a two step strategy. Here, we use Newton’s second law to determine the acceleration of the suitcase and then use kinematics to determine the suitcase’s speed after lifting 0.50 s.
Simplify and diagram Next, we construct a force diagram for the suitcase while being lifted. The y-components of the forces exerted on the suitcase are your upward pull on the suitcase F_{\mathrm{Y} \text { on } S y}=+F_{\mathrm{Y} \text { on } \mathrm{S}}=+120 \mathrm{~N} a n d Earth’s downward pull on the suitcase F_{\mathrm{E} \text { on } \mathrm{S} y}=-F_{\mathrm{E} \text { on } \mathrm{S}}=-100 \mathrm{~N} . Because the upward force is larger, the suitcase will have an upward acceleration \vec{a} .
Represent mathematically Since all the forces are along the y-axis, we apply the y-component form of Newton’s second law to determine the suitcase’s acceleration (notice how the subscripts in the equation below change from step to step):
a_{\mathrm{S} y}=\frac{\Sigma F_{\text {on } \mathrm{S} y}}{m_{\mathrm{S}}}=\frac{F_{\mathrm{Y} \text { on } \mathrm{S} y}+F_{\mathrm{E} \text { on } \mathrm{S} y}}{m_{\mathrm{S}}}
=\frac{\left(+F_{\mathrm{Y} \text { on } \mathrm{S}}\right)+\left(-F_{\mathrm{E} \text { on } \mathrm{S}}\right)}{m_{\mathrm{S}}}
=\frac{F_{\mathrm{Y} \text { on } \mathrm{S}}-F_{\mathrm{E} \text { on } \mathrm{S}}}{m_{\mathrm{S}}}
After using Newton’s second law to determine the acceleration of the suitcase, we then use kinematics to determine the suitcase’s speed after traveling upward for 0.50 s:
v_y=v_{0 y}+a_y t
The initial velocity is v_{0 y}=0
Solve and evaluate
Now substitute the known information in the Newton’s second law y-component equation above to find the acceleration of the suitcase:
a_{\mathrm{S} y}=\frac{F_{\mathrm{Y} \text { on } \mathrm{S}}-F_{\mathrm{E} \text { on } \mathrm{S}}}{m_{\mathrm{S}}}=\frac{120 \mathrm{~N}-100 \mathrm{~N}}{10 \mathrm{~kg}}=+2.0 \mathrm{~m} / \mathrm{s}^2
Insert this and other known information into the kinematics equation to find the vertical velocity of the suitcase after lifting for 0.50 s:
v_y=v_{0 y}+a_y t=0+\left(+2.0 \mathrm{~m} / \mathrm{s}^2\right)(0.50 \mathrm{~s})=+1.0 \mathrm{~m} / \mathrm{s}
The unit for time is correct and the magnitude is reasonable.
Try it yourself: How far up did you pull the suitcase during this 0.50 s?
Answer : The average speed while lifting it was (0 + 1.0 m/s)/2 = 0.50 m/s. Thus you lifted the suitcase y-y_0=(0.50 \mathrm{~m} / \mathrm{s})(0.50 \mathrm{~s})=0.25 \mathrm{~m}