A person jumps from the roof of a house 3.8-m high When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.73 m Part A Part B If the mass of his forso (excluding legs) is 42 kg , find the magnitude of the average force exerted on his torso by his legs during deceleration Express your answer using two significant figures. F=

To find the magnitude of the average force exerted on the torso by the legs during deceleration, we can use the equation of motion:<br /><br />\[ v^2 = u^2 + 2as \]<br /><br />where:<br />- \( v \) is the final velocity (0 m/s since he comes to rest),<br />- \( u \) is the initial velocity,<br />- \( a \) is the acceleration (deceleration in this case),<br />- \( s \) is the distance over which the deceleration occurs.<br /><br />First, we need to find the initial velocity \( u \) when he starts to decelerate. We can use the equation:<br /><br />\[ v^2 = u^2 + 2as \]<br /><br />Rearranging for \( u \):<br /><br />\[ u^2 = v^2 - 2as \]<br /><br />Given:<br />- \( v = 0 \) m/s (final velocity),<br />- \( a = -g \) (acceleration due to gravity, negative because it's deceleration),<br />- \( s = 0.73 \) m (distance of deceleration).<br /><br />Substituting the values:<br /><br />\[ 0 = u^2 - 2(-9.8 \, \text{m/s}^2)(0.73 \, \text{m}) \]<br /><br />\[ u^2 = 2(9.8 \, \text{m/s}^2)(0.73 \, \text{m}) \]<br /><br />\[ u^2 = 14.224 \, \text{m}^2/\text{s}^2 \]<br /><br />\[ u = \sqrt{14.224} \]<br /><br />\[ u \approx 3.77 \, \text{m/s} \]<br /><br />Now, we can find the magnitude of the average force exerted on the torso by the legs during deceleration using the equation:<br /><br />\[ F = ma \]<br /><br />where:<br />- \( m = 42 \, \text{kg} \) (mass of the torso),<br />- \( a = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it's deceleration).<br /><br />Substituting the values:<br /><br />\[ F = 42 \, \text{kg} \times 9.8 \, \text{m/s}^2 \]<br /><br />\[ F = 411.6 \, \text{N} \]<br /><br />Therefore, the magnitude of the average force exerted on the torso by the legs during deceleration is approximately \( 410 \, \text{N} \) (rounded to two significant figures).