Search any question & find its solution
Question:
Answered & Verified by Expert
A pebble of mass $0.05 \mathrm{~kg}$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point, where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^{\circ}$ with the horizontal direction? Ignore air resistance.
Solution:
2042 Upvotes
Verified Answer
(a) Ignore air resistance. When the pebble is moving upward, the acceleration $\mathrm{g}$ is acting downward, so the force is acting downward is equal to $F=m g=0.05 \mathrm{~kg} \times 10 \mathrm{~ms}^{-2}=0.5 \mathrm{~N}$.
(b) In this case also $F=m g=0.05 \times 10=0.5$ N. (downwards).
(c) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at $45^{\circ}$ because no other acceleration except ' $g$ ' is acting on pebble.
(b) In this case also $F=m g=0.05 \times 10=0.5$ N. (downwards).
(c) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at $45^{\circ}$ because no other acceleration except ' $g$ ' is acting on pebble.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.