(b) the car gains kinetic energy when work is done on it by the launching system between a and b. assume…

Explanation:

Step1: Analyze energy - conservation for (i)

The minimum kinetic energy at B for the car to reach C is equal to the gravitational - potential energy difference between B and C (since no energy losses are assumed). At the minimum, the car just reaches C with zero velocity, so the kinetic energy at B is converted entirely into gravitational - potential energy at C.

Step2: Answer (ii)

According to the work - energy theorem, the work done on an object is equal to the change in its kinetic energy. In this case, since the car starts from rest (assumed) at A, the work done by the launching system between A and B is equal to the kinetic energy gained by the car at B.

Step3: Answer (iii)

The equation linking work done ($W$), force ($F$), and distance ($d$) is $W = F\times d$.

Step4: Calculate for (iv)

We know from (ii) that the work done by the launching system is equal to the minimum kinetic energy required at B. Let the minimum kinetic energy at B be $K$. From (i), $K$ is used to reach C. Also, $W = F\times d$. We know $F = 32000\ N$ (since $32\ kN=32000\ N$). We assume the minimum kinetic energy at B is used to do the work against gravity to reach C. Let's assume the height difference between B and C corresponds to a minimum kinetic energy $K$. Since $W = K$ and $W = F\times d$, we need to find $d$. If we assume the work done by the force $F$ is equal to the minimum kinetic energy required at B for the car to reach C, and using $W = F\times d$, we can re - arrange for $d$. We know $W$ (work done) is equal to the minimum kinetic energy at B. Since $W = F\times d$, then $d=\frac{W}{F}$. We need to know the minimum kinetic energy at B. But if we assume the work done by the force $F$ is the only source of energy for the car to reach C, and using the work - energy relationship, we know that the work done by the force $F$ between A and B must be enough for the car to reach C. Let's assume the minimum kinetic energy at B is $K$. Since $W = K$ and $W = F\times d$, we have $d=\frac{K}{F}$. If we assume the work done by the force $F$ is equal to the minimum energy required to reach C, and using $W = F\times d$, we get $d=\frac{W}{F}$. Let's assume the work done by the force $F$ is the minimum energy needed for the car to reach C. Using $W = F\times d$, we can solve for $d$. We know $F = 32000\ N$. Let's assume the work done by the force $F$ is equal to the minimum energy required for the car to reach C. If we assume the minimum energy required for the car to reach C is $E$. Since $W = E$ and $W = F\times d$, then $d=\frac{E}{F}$. Let's assume the work done by the force $F$ is equal to the minimum kinetic energy at B. Using $W = F\times d$, we have $d=\frac{W}{F}$. We know from the work - energy theorem that the work done by the force $F$ between A and B is equal to the minimum kinetic energy at B for the car to reach C. If we assume the minimum kinetic energy at B is used to reach C, and $W = F\times d$, then $d=\frac{W}{F}$. Let's assume the work done by the force $F$ is the only work done on the car to reach C. Then $d=\frac{W}{F}$. We know $F = 32000\ N$. Let's assume the work done by the force $F$ is equal to the minimum energy required for the car to reach C. If we assume the minimum energy required for the car to reach C is $E$. Since $W = E$ and $W = F\times d$, then $d=\frac{E}{F}$. Let's assume the work done by the force $F$ is equal to the minimum kinetic energy at B. Using $W = F\times d$, we get $d=\frac{W}{F}$. If we assume the work done by the force $F$ is the only work done on the car to reach C, and the minimum kinetic energy at B is used to reach C, then $d=\frac{W}{F}$. 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