Paragraph: A small block of mass $1 \mathrm{~kg}$ is released from rest at the top of a rough track. The track is a circular arc of radius $40 \mathrm{~m}$. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity.…

Question

Paragraph: A small block of mass $1 \mathrm{~kg}$ is released from rest at the top of a rough track. The track is a circular arc of radius $40 \mathrm{~m}$. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point $Q$, as shown in the figure below, is $150 \mathrm{~J}$. (Take the acceleration due to gravity, $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ ). Question: The speed of the block when it reaches the point $Q$ is

Aicharya · Accepted Answer

Apply conservation of energy, &#8710; P E - W f = &#8710; K E M g R sin 30 &#176; - 150 = 1 2 M v 2 M g R 2 - 150 = 1 2 M v 2 &#8658; 1 &#215; 10 &#215; 20 - 150 = 1 2 &#215; 1 &#215; v 2 &#8658; v = 10 &#160; m &#160; s - 1 .

Paragraph: A small block of mass $1 \mathrm{~kg}$ is released from rest at the top of a rough track. The…

Solution