Question

When a horizontal force P acts on a cart of mass 20 kg, it moves with a uniform velocity on a horizontal floor. When a force of 1.2P acts on the cart, it moves with an acceleration of 0.05 m/st. Find the value of P.​

Answer 1

$\bf \blue{Given \: data:} \\ \sf { Mass \: of \: the \: cart, \: m = 20 \: kg} \\ \sf {P =?} \\\sf{ The \: horizontal \: floor \: will \: have \: a \: frictional \: force \: F \: that \: obstructs \: the \: movement \: of \: the \: cart. }\\ \sf{Thus \: acceleration \: of \: the \: cart, \: a = P - F} \\\sf{Since \: it \: is \: moving \: with \: a \: uniform \: velocity, the \: acceleration \: will \: be \: zero.} \\ \sf{P - F = 0 }\\\sf{ P = F} \\\sf{ Hence \: when \: a \: force \: of \: 1.2P \: acts \: on \: the \: cart, \: the \: acceleration \: of \: the \: cart \: is \: 1.2P - F = ma }\\ \sf{= 1.2P-P = 20 \times 0.05} \\\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:( a = 0.05ms { - }^{2} )} \\ \sf{ = P = \frac{1}{0.2} = 5N} \\ \sf{Thus \: the \: value \: of \: P \: is} \: \blue{ 5N}$