Kvant Physics Problem 197

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Problem

On a body of mass $m$, lying on a horizontal rough surface with a coefficient of friction $k$, at time $t=0$ a force proportional to time began to act at an angle $\alpha$ to the horizontal. Determine the velocity $v$ of the body after $\tau$ seconds.

A. V. Ustinova

Setup and Assumptions

A body of mass $m$ lies on a horizontal rough surface with coefficient of kinetic friction $k$. At time $t=0$ a force whose magnitude grows linearly with time acts on the body. The force is written as $F(t)=\lambda t$, where $\lambda$ has units $\mathrm{N,s^{-1}}$. The force is applied at a constant angle $\alpha$ to the horizontal and is directed so that its horizontal component drives the motion.

The motion is considered in an inertial frame attached to the ground. Air resistance is neglected. The coefficient of friction is taken as constant and independent of speed. The body is assumed to slide from the beginning of motion so that kinetic friction applies for all $t>0$. The normal reaction adjusts instantaneously. The initial velocity at $t=0$ is zero. The angle $\alpha$ is measured from the horizontal toward the direction that has an upward vertical component, so the vertical component of the applied force reduces the normal force.

The unknown quantity is the speed $v(\tau)$ after time $\tau$.

Physical Principles

Newton’s second law is applied separately along horizontal and vertical directions in an inertial frame, written as

$m\frac{dv_x}{dt}=\sum F_x,$

and the normal force is determined from vertical force balance in the absence of vertical acceleration,

$\sum F_y = 0.$

The kinetic friction force has magnitude

$F_f = kN,$

and acts opposite to the direction of motion.

The applied force components are

$F_x(t)=\lambda t\cos\alpha, \quad F_y(t)=\lambda t\sin\alpha.$

Derivation

Vertical motion is constrained, so the net vertical force vanishes. Taking upward as positive, the vertical force balance reads

$N + \lambda t\sin\alpha - mg = 0,$

which gives the normal force

$N = mg - \lambda t\sin\alpha.$

The kinetic friction force is therefore

$F_f = k\left(mg - \lambda t\sin\alpha\right).$

The horizontal equation of motion is

$m\frac{dv}{dt} = \lambda t\cos\alpha - k\left(mg - \lambda t\sin\alpha\right).$

Expanding the right-hand side yields

$m\frac{dv}{dt} = \lambda t\cos\alpha + k\lambda t\sin\alpha - kmg.$

Grouping the terms proportional to time gives

$m\frac{dv}{dt} = \lambda t(\cos\alpha + k\sin\alpha) - kmg.$

Dividing by $m$ produces

$\frac{dv}{dt} = \frac{\lambda}{m}t(\cos\alpha + k\sin\alpha) - kg.$

Integrating from $t=0$ to $t=\tau$ with $v(0)=0$ gives

$v(\tau)=\frac{\lambda}{m}(\cos\alpha + k\sin\alpha)\int_0^\tau t,dt - kg\int_0^\tau dt.$

Evaluating the integrals,

$\int_0^\tau t,dt = \frac{\tau^2}{2}, \quad \int_0^\tau dt = \tau,$

so the velocity becomes

$v(\tau)=\frac{\lambda}{m}(\cos\alpha + k\sin\alpha)\frac{\tau^2}{2} - kg\tau.$

Result

The velocity after time $\tau$ is

$v(\tau)=\frac{\lambda \tau^2}{2m}\left(\cos\alpha + k\sin\alpha\right) - kg\tau.$

No numerical values are provided in the statement, so the expression already represents the final quantitative result in SI base form. The parameters carry units $\lambda$ in $\mathrm{N,s^{-1}}$, $\tau$ in $\mathrm{s}$, $m$ in $\mathrm{kg}$, and $g$ in $\mathrm{m,s^{-2}}$, ensuring $v$ has units $\mathrm{m,s^{-1}}$.

Sanity Checks

Dimensional consistency follows from $\lambda \tau^2/m$, which has units $(\mathrm{N,s^{-1}})\mathrm{s^2}/\mathrm{kg}=\mathrm{N,s}/\mathrm{kg}=\mathrm{m,s^{-1}}$, matching the velocity unit. The term $kg\tau$ has units $(1)\mathrm{m,s^{-2}}\mathrm{s}=\mathrm{m,s^{-1}}$, consistent with subtraction.

In the limit $\lambda \to 0$, the expression reduces to $v(\tau)=-kg\tau$, corresponding to deceleration under friction alone with no driving force. In the limit $k \to 0$, friction disappears and the result becomes $v(\tau)=\lambda \tau^2\cos\alpha/(2m)$, corresponding to motion driven solely by the horizontal component of a linearly increasing force. A sign error most readily arises in the normal force expression, since reversing the sign of the vertical force component would incorrectly increase friction instead of decreasing it.