Answer
-6.75 J
Work Step by Step
$F=mg=1.50\,kg\times9.8\,m/s^{2}=14.7\,N$
d= maximum height of the projectile=$\frac{(v_{0}\sin\theta_{0})^{2}}{2g}$
$=\frac{(3.00\,m/s\times\sin 90^{\circ})^{2}}{2\times9.8\,m/s^{2}}=0.459\,m$
Angle between force vector and displacement vector $\theta=180^{\circ}$
$W=Fd\cos\theta=14.7\,N\times0.459\,m\times\cos180^{\circ}=-6.75\,J$
The other method to solve this problem:
$K_{1}=6.75\,J$ (from the previous section)
$U_{1}=mgh_{1}=mg\times0=0$
$K_{2}=\frac{1}{2}mv_{f}^{2}=\frac{1}{2}m\times(0)^{2}=0$
$U_{2}=mgh_{2}$
According to the principle of conservation of mechanical energy,
$K_{2}+U_{2}=K_{1}+U_{1}$
Therefore, we have
$mgh_{2}=6.75\,J$
Now, $W=Fd\cos\theta=(mg)(h_{2})\cos180^{\circ}=-mgh_{2}=-6.75\,J$
