Write a realistic problem for which this is the correct equation(s).
$T-\left(1500\text{kg}\right)\left(9.8{\text{m/s}}^2\right)=\left(1500\text{kg}\right)\left(1.0{\text{m/s}}^2\right)$
$P=T\left(2.0\text{m/s}\right)$

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. How long does it take the Porsche to reach the maximum power output?

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. If the Porsche accelerates at a_max, what is its speed when it reaches maximum power output?

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. What is the maximum acceleration of the Porsche on a concrete surface where μ_s = 1.00? Hint: What force pushes the car forward?

A 150 g particle at x = 0 is moving at 2.00 m/s in the + x - direction. As it moves, it experiences a force given by Fₓ = (0.250 N) sin (x/2.00 m). What is the particle's speed when it reaches x = 3.14 m?

A particle moving on the x-axis experiences a force given by F_x = qx², where q is a constant. How much work is done on the particle as it moves from x = 0 to x = d?

Draw a pictorial representation.

$T-(1500\,\text{kg})(9.8\,\text{m/s}^2)=(1500\,\text{kg})\left(1.0\,\text{m/s}^2\right)$

$P=T(2.0\,\text{m/s})$

Finish the solution of the problem.

$T-(1500\,\text{kg})(9.8\,\text{m/s}^2)=(1500\,\text{kg})\left(1.0\,\text{m/s}^2\right)$

$P=T(2.0\,\text{m/s})$

The gravitational attraction between two objects with masses m_A and m_B, separated by distance 𝓍, is F = Gm_Am_B/𝓍², where G is the gravitational constant. If one mass is much greater than the other, the larger mass stays essentially at rest while the smaller mass moves toward it. Suppose a 1.5 x 10¹³ kg comet is passing the orbit of Mars, heading straight for the sun at a speed of 3.5 x 10⁴ m/s. What will its speed be when it crosses the orbit of Mercury? Astronomical data are given in the tables at the back of the book, and G = 6.67 x 10^-11 Nm²/kg².